CÁC QUY TẮC TÍNH ĐẠO HÀM
A – TÓM TẮT LÝ THUYẾT
1. Quy tắc tính đạo hàm
(C)' = 0
(x)' = 1
$({x^n})' = n{x^{n - 1}},n \in {\mathbb{N}^*}$
\({\left( {\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt x }}\)
2. Đạo hàm của tổng, hiệu, tích, thương của hàm số
\( \bullet \)${({\rm{u}} \pm {\rm{v}})^\prime } = {\rm{ }}{{\rm{u}}^\prime } \pm {{\rm{v}}^\prime }$$ \Rightarrow {\rm{ }}({u_1} \pm {u_2} \pm ... \pm {u_n})' = u_1^, \pm u_2^, \pm ... \pm u_n^,$
\( \bullet \)${({\rm{uv}})^\prime } = {\rm{ }}{{\rm{u}}^\prime }{\rm{v }} + {\rm{ }}{{\rm{v}}^\prime }{\rm{u}}$$ \Rightarrow (uvw)' = u'vw + uv'w + uvw'$
${({\rm{ku}})^\prime } = {\rm{ k}}{{\rm{u}}^\prime }$
\( \bullet \)\({\left( {\frac{u}{v}} \right)^\prime } = \frac{{u\prime v - v\prime u}}{{{v^2}}}\) \( \Rightarrow {\left( {\frac{1}{v}} \right)^\prime } = - \frac{{v\prime }}{{{v^2}}}\).
3. Đạo hàm của hàm số hợp
Cho hàm số $y = f(u(x)) = f(u)$với$u = u(x)$. Khi đó$y{'_x} = y{'_u}.u{'_x}$.
4. Bảng công thức đạo hàm các hàm sơ cấp cơ bản
(c)' = 0
(x)' = 1
$({x^\alpha })' = \alpha {x^{\alpha - 1}}$
$\left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }}$
$\left( {\sqrt[n]{x}} \right)' = \frac{1}{{n\sqrt[n]{{{x^{n - 1}}}}}}$
$\left( {{u^\alpha }} \right)' = \alpha {u^{\alpha - 1}}.u'$
$\left( {\sqrt u } \right)' = \frac{{u'}}{{2\sqrt u }}$
$\left( {\sqrt[n]{u}} \right)' = \frac{{u'}}{{n\sqrt[n]{{{u^{n - 1}}}}}}$
A – TÓM TẮT LÝ THUYẾT
1. Quy tắc tính đạo hàm
(C)' = 0
(x)' = 1
$({x^n})' = n{x^{n - 1}},n \in {\mathbb{N}^*}$
\({\left( {\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt x }}\)
2. Đạo hàm của tổng, hiệu, tích, thương của hàm số
\( \bullet \)${({\rm{u}} \pm {\rm{v}})^\prime } = {\rm{ }}{{\rm{u}}^\prime } \pm {{\rm{v}}^\prime }$$ \Rightarrow {\rm{ }}({u_1} \pm {u_2} \pm ... \pm {u_n})' = u_1^, \pm u_2^, \pm ... \pm u_n^,$
\( \bullet \)${({\rm{uv}})^\prime } = {\rm{ }}{{\rm{u}}^\prime }{\rm{v }} + {\rm{ }}{{\rm{v}}^\prime }{\rm{u}}$$ \Rightarrow (uvw)' = u'vw + uv'w + uvw'$
${({\rm{ku}})^\prime } = {\rm{ k}}{{\rm{u}}^\prime }$
\( \bullet \)\({\left( {\frac{u}{v}} \right)^\prime } = \frac{{u\prime v - v\prime u}}{{{v^2}}}\) \( \Rightarrow {\left( {\frac{1}{v}} \right)^\prime } = - \frac{{v\prime }}{{{v^2}}}\).
3. Đạo hàm của hàm số hợp
Cho hàm số $y = f(u(x)) = f(u)$với$u = u(x)$. Khi đó$y{'_x} = y{'_u}.u{'_x}$.
4. Bảng công thức đạo hàm các hàm sơ cấp cơ bản
(c)' = 0
(x)' = 1
$({x^\alpha })' = \alpha {x^{\alpha - 1}}$
$\left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }}$
$\left( {\sqrt[n]{x}} \right)' = \frac{1}{{n\sqrt[n]{{{x^{n - 1}}}}}}$
$\left( {{u^\alpha }} \right)' = \alpha {u^{\alpha - 1}}.u'$
$\left( {\sqrt u } \right)' = \frac{{u'}}{{2\sqrt u }}$
$\left( {\sqrt[n]{u}} \right)' = \frac{{u'}}{{n\sqrt[n]{{{u^{n - 1}}}}}}$
B – BÀI TẬP
Kiểu 1: TÍNH ĐẠO HÀN BẰNG CÔNG THỨC TẠI MỘT ĐIỂM HOẶC BẰNG MÁY TÍNH CẦM TAY
Câu 1. Cho hàm số f(x) xác định trên \(\mathbb{R}\) bởi \(f\left( x \right) = 2{x^2} + 1\). Giá trị \(f'\left( { - 1} \right)\) bằng:
A. 2.
B. 6.
C. - 4.
D. 3.
Chọn C
Ta có : \(f'\left( x \right) = 4x\) \( \Rightarrow f'\left( { - 1} \right) = - 4\).
Ta có : \(f'\left( x \right) = 4x\) \( \Rightarrow f'\left( { - 1} \right) = - 4\).
A. 4.
B. 14.
C. 15.
D. 24.
Chọn D
·Ta có: \(f'\left( x \right)\)$ = - 4{x^3} + 12{x^2} - 6x + 2$. Nên $f'\left( { - 1} \right)$$ = 24$.
·Ta có: \(f'\left( x \right)\)$ = - 4{x^3} + 12{x^2} - 6x + 2$. Nên $f'\left( { - 1} \right)$$ = 24$.
A. - 32.
B. 30.
C. - 64.
D. 12.
Chọn C
Ta có : \(y' = 4{\left( {{x^2} + 1} \right)^3}{\left( {{x^2} + 1} \right)^\prime } = 8x{\left( {{x^2} + 1} \right)^3}\)
\( \Rightarrow y'\left( { - 1} \right) = - 64\).
Ta có : \(y' = 4{\left( {{x^2} + 1} \right)^3}{\left( {{x^2} + 1} \right)^\prime } = 8x{\left( {{x^2} + 1} \right)^3}\)
\( \Rightarrow y'\left( { - 1} \right) = - 64\).
A. 1.
B. - 3.
C. - 5.
D. 0.
Chọn D
Ta có: \(f(x) = \frac{{{x^2} - 2x + 5}}{{x - 1}}\)$ = x - 1 + \frac{4}{{x - 1}}$$ \Rightarrow f'\left( x \right) = 1 - \frac{4}{{{{\left( {x - 1} \right)}^2}}}$$ \Rightarrow f'\left( { - 1} \right) = 0$.
Ta có: \(f(x) = \frac{{{x^2} - 2x + 5}}{{x - 1}}\)$ = x - 1 + \frac{4}{{x - 1}}$$ \Rightarrow f'\left( x \right) = 1 - \frac{4}{{{{\left( {x - 1} \right)}^2}}}$$ \Rightarrow f'\left( { - 1} \right) = 0$.
A. 0.
B. 2.
C. 1.
D. Không tồn tại.
Chọn D
Ta có : \(f'\left( x \right) = \frac{x}{{\sqrt {{x^2}} }}\)
\( \Rightarrow f'\left( x \right)\) không xác định tại x = 0
\( \Rightarrow f'\left( 0 \right)\) không có đạo hàm tại x = 0.
Ta có : \(f'\left( x \right) = \frac{x}{{\sqrt {{x^2}} }}\)
\( \Rightarrow f'\left( x \right)\) không xác định tại x = 0
\( \Rightarrow f'\left( 0 \right)\) không có đạo hàm tại x = 0.
A. \(y'\left( 0 \right) = \frac{1}{2}\).
B. \(y'\left( 0 \right) = \frac{1}{3}\).
C. \(y'\left( 0 \right) = 1\).
D. \(y'\left( 0 \right) = 2\).
Chọn A
Ta có : \(y' = \frac{{\sqrt {4 - {x^2}} - x\frac{{ - x}}{{\sqrt {4 - {x^2}} }}}}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^2}}} = \frac{4}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^3}}}\)
\( \Rightarrow y'\left( 0 \right) = \frac{1}{2}\).
Ta có : \(y' = \frac{{\sqrt {4 - {x^2}} - x\frac{{ - x}}{{\sqrt {4 - {x^2}} }}}}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^2}}} = \frac{4}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^3}}}\)
\( \Rightarrow y'\left( 0 \right) = \frac{1}{2}\).
A. \(\frac{1}{{12}}\).
B. \( - \frac{1}{{12}}\).
C. \(\frac{1}{6}\).
D. \( - \frac{1}{6}\).
Chọn A
Ta có : \(y = \sqrt[3]{x} \Rightarrow {y^3} = x \Rightarrow 3{y^2}.y' = 1 \Rightarrow y' = \frac{1}{{3{y^2}}} = \frac{1}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}}\)
\( \Rightarrow y'\left( { - 8} \right) = \frac{1}{{12}}\).
Ta có : \(y = \sqrt[3]{x} \Rightarrow {y^3} = x \Rightarrow 3{y^2}.y' = 1 \Rightarrow y' = \frac{1}{{3{y^2}}} = \frac{1}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}}\)
\( \Rightarrow y'\left( { - 8} \right) = \frac{1}{{12}}\).
A. 1/2.
B. \( - \frac{1}{2}\).
C. - 2.
D. Không tồn tại.
Chọn B
Ta có : \(f'\left( x \right) = \frac{{2\left( {x - 1} \right) - 2x}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}\) \( \Rightarrow f'\left( { - 1} \right) = - \frac{1}{2}\).
Ta có : \(f'\left( x \right) = \frac{{2\left( {x - 1} \right) - 2x}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}\) \( \Rightarrow f'\left( { - 1} \right) = - \frac{1}{2}\).
A. 0.
B. 1.
C. 1/2.
D. Không tồn tại.
Chọn C
Ta có : \(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1}}{{{x^2}}}\) \( = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 1} + 1}} = \frac{1}{2}\).
Ta có : \(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1}}{{{x^2}}}\) \( = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {{x^2} + 1} + 1}} = \frac{1}{2}\).
A. \(y'\left( 1 \right) = - 4\).
B. \(y'\left( 1 \right) = - 5\).
C. \(y'\left( 1 \right) = - 3\).
D. \(y'\left( 1 \right) = - 2\).
Chọn B
Ta có : \(y' = \frac{{\left( {2x + 1} \right)\left( {x - 2} \right) - \left( {{x^2} + x} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x - 2}}{{{{\left( {x - 2} \right)}^2}}}\)
\( \Rightarrow y'\left( 1 \right) = - 5\).
Ta có : \(y' = \frac{{\left( {2x + 1} \right)\left( {x - 2} \right) - \left( {{x^2} + x} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x - 2}}{{{{\left( {x - 2} \right)}^2}}}\)
\( \Rightarrow y'\left( 1 \right) = - 5\).
A. \(y'\left( 0 \right) = \frac{1}{2}\).
B. \(y'\left( 0 \right) = \frac{1}{3}\).
C. \(y'\left( 0 \right) = 1\).
D. \(y'\left( 0 \right) = 2\).
Chọn A
$\begin{array}{l} y' = f'(x) = {\left( {\frac{x}{{\sqrt {4 - {x^2}} }}} \right)^,} = \frac{{x'.\sqrt {4 - {x^2}} - x.{{\left( {\sqrt {4 - {x^2}} } \right)}^,}}}{{4 - {x^2}}} = \frac{{\sqrt {4 - {x^2}} + \frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}}}{{4 - {x^2}}}\\ \Rightarrow y'\left( 0 \right) = \frac{{\sqrt 4 }}{4} = \frac{1}{2} \end{array}$
$\begin{array}{l} y' = f'(x) = {\left( {\frac{x}{{\sqrt {4 - {x^2}} }}} \right)^,} = \frac{{x'.\sqrt {4 - {x^2}} - x.{{\left( {\sqrt {4 - {x^2}} } \right)}^,}}}{{4 - {x^2}}} = \frac{{\sqrt {4 - {x^2}} + \frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}}}{{4 - {x^2}}}\\ \Rightarrow y'\left( 0 \right) = \frac{{\sqrt 4 }}{4} = \frac{1}{2} \end{array}$
A. $y'\left( 1 \right) = - 4$.
B. $y'\left( 1 \right) = - 3$.
C. $y'\left( 1 \right) = - 2$.
D. $y'\left( 1 \right) = - 5$.
Chọn D
Ta có: $y = \frac{{{x^2} + x}}{{x - 2}}$$ = x + 3 + \frac{6}{{x - 2}}$$ \Rightarrow y' = 1 - \frac{6}{{{{\left( {x - 2} \right)}^2}}}$$ \Rightarrow y'\left( 1 \right) = 1 - 6 = - 5$.
Ta có: $y = \frac{{{x^2} + x}}{{x - 2}}$$ = x + 3 + \frac{6}{{x - 2}}$$ \Rightarrow y' = 1 - \frac{6}{{{{\left( {x - 2} \right)}^2}}}$$ \Rightarrow y'\left( 1 \right) = 1 - 6 = - 5$.
A. $\frac{1}{6}$.
B. $\frac{1}{{12}}$.
C. -$\frac{1}{6}$.
D. $ - \frac{1}{{12}}$.
:
Với $x > 0$
${f^\prime }\left( x \right)\, = {\left( {{x^{\frac{1}{3}}}} \right)^\prime } = \frac{1}{3}{x^{\frac{{ - 2}}{3}}} \Rightarrow {f^\prime }\left( 8 \right) = \frac{1}{3}{.8^{\frac{{ - 2}}{3}}} = \frac{1}{3}{2^{ - 2}} = \frac{1}{{12}}$.
Đáp án
B.
Với $x > 0$
${f^\prime }\left( x \right)\, = {\left( {{x^{\frac{1}{3}}}} \right)^\prime } = \frac{1}{3}{x^{\frac{{ - 2}}{3}}} \Rightarrow {f^\prime }\left( 8 \right) = \frac{1}{3}{.8^{\frac{{ - 2}}{3}}} = \frac{1}{3}{2^{ - 2}} = \frac{1}{{12}}$.
Đáp án
B.
A. 1/2.
B. 1.
C. 0
D. Không tồn tại.
Đáp án
D.
Ta có $f'\left( x \right) = \frac{1}{{2\sqrt {x - 1} }}$
D.
Ta có $f'\left( x \right) = \frac{1}{{2\sqrt {x - 1} }}$
A. Không xác định.
B. -3
C. 3
D. 0
Hàm số không xác định tại nên không xác định
Chọn A
Chọn A
A. -14
B. 12
C. 13
D. 10
Chọn A
Bước đầu tiên tính đạo hàm sử dụng công thức ${\left( {\frac{1}{{{x^\alpha }}}} \right)^/} = \frac{{ - \alpha }}{{{x^{\alpha + 1}}}}$
$f'\left( x \right) = {\left( {\frac{1}{x} + \frac{2}{{{x^2}}} + \frac{3}{{{x^3}}}} \right)^/} = - \frac{1}{{{x^2}}} - \frac{4}{{{x^3}}} - \frac{9}{{{x^4}}}$$ \Rightarrow f'\left( 1 \right) = - 1 - 4 - 9 = - 14$
Bước đầu tiên tính đạo hàm sử dụng công thức ${\left( {\frac{1}{{{x^\alpha }}}} \right)^/} = \frac{{ - \alpha }}{{{x^{\alpha + 1}}}}$
$f'\left( x \right) = {\left( {\frac{1}{x} + \frac{2}{{{x^2}}} + \frac{3}{{{x^3}}}} \right)^/} = - \frac{1}{{{x^2}}} - \frac{4}{{{x^3}}} - \frac{9}{{{x^4}}}$$ \Rightarrow f'\left( 1 \right) = - 1 - 4 - 9 = - 14$
A. 1/2
B. 1
C. 2
D. 3
Chọn A
Ta có $f'\left( x \right) = {\left( {\frac{1}{x} + \frac{1}{{\sqrt x }} + {x^2}} \right)^/} = - \frac{1}{{{x^2}}} - \frac{{{{\left( {\sqrt x } \right)}^/}}}{x} + 2x = - \frac{1}{{{x^2}}} - \frac{1}{{2x\sqrt x }} + 2x$
Vậy $f'\left( 1 \right) = - 1 - \frac{1}{2} + 2 = \frac{1}{2}$
Ta có $f'\left( x \right) = {\left( {\frac{1}{x} + \frac{1}{{\sqrt x }} + {x^2}} \right)^/} = - \frac{1}{{{x^2}}} - \frac{{{{\left( {\sqrt x } \right)}^/}}}{x} + 2x = - \frac{1}{{{x^2}}} - \frac{1}{{2x\sqrt x }} + 2x$
Vậy $f'\left( 1 \right) = - 1 - \frac{1}{2} + 2 = \frac{1}{2}$
A. 4
B. 5
C. 6
D. 7
Chọn A
Ta có $f'\left( x \right) = {\left( {{x^5} + {x^3} - 2x - 3} \right)^/} = 5{x^4} + 3{x^2} - 2$
$f'\left( 1 \right) + f'\left( { - 1} \right) + 4f\left( 0 \right) = (5 + 3 - 2) + (5 + 3 - 2) + 4.( - 2) = 4$
Ta có $f'\left( x \right) = {\left( {{x^5} + {x^3} - 2x - 3} \right)^/} = 5{x^4} + 3{x^2} - 2$
$f'\left( 1 \right) + f'\left( { - 1} \right) + 4f\left( 0 \right) = (5 + 3 - 2) + (5 + 3 - 2) + 4.( - 2) = 4$
A. \(\frac{1}{4}\)
B. 1
C. 2
D. 3
Chọn A
$f'\left( x \right) = {\left( {\frac{x}{{\sqrt {4 - {x^2}} }}} \right)^/} = \frac{{x'\sqrt {4 - {x^2}} - x{{\left( {\sqrt {4 - {x^2}} } \right)}^/}}}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^2}}} = \frac{{\sqrt {4 - {x^2}} + \frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}}}{{\left( {4 - {x^2}} \right)}} = \frac{4}{{\left( {4 - {x^2}} \right)\sqrt {4 - {x^2}} }}$
Vậy $f'\left( 0 \right) = \frac{1}{4}$.
$f'\left( x \right) = {\left( {\frac{x}{{\sqrt {4 - {x^2}} }}} \right)^/} = \frac{{x'\sqrt {4 - {x^2}} - x{{\left( {\sqrt {4 - {x^2}} } \right)}^/}}}{{{{\left( {\sqrt {4 - {x^2}} } \right)}^2}}} = \frac{{\sqrt {4 - {x^2}} + \frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}}}{{\left( {4 - {x^2}} \right)}} = \frac{4}{{\left( {4 - {x^2}} \right)\sqrt {4 - {x^2}} }}$
Vậy $f'\left( 0 \right) = \frac{1}{4}$.
A. \( - \frac{{11}}{3}.\)
B. \(\frac{1}{5}.\)
C. $ - {\rm{11}}.$
D. \( - \frac{{11}}{9}.\)
Chọn C
\(f'\left( x \right) = \frac{{ - 11}}{{{{\left( {2x + 1} \right)}^2}}} \Rightarrow f'\left( { - 1} \right) = \frac{{ - 11}}{1} = - 11\).
\(f'\left( x \right) = \frac{{ - 11}}{{{{\left( {2x + 1} \right)}^2}}} \Rightarrow f'\left( { - 1} \right) = \frac{{ - 11}}{1} = - 11\).
A. $ - \frac{5}{8}.$
B. $\frac{{25}}{{16}}.$
C. $\frac{5}{8}.$
D. $\frac{{11}}{8}.$
Chọn C
\(f'\left( x \right) = \frac{{ - 6}}{{{{\left( {x + 3} \right)}^2}}} + \frac{2}{{\sqrt {4x} }}\)
\(f'\left( 1 \right) = \frac{{ - 6}}{{{{\left( {1 + 3} \right)}^2}}} + \frac{2}{{\sqrt {4.1} }} = \frac{5}{8}\).
\(f'\left( x \right) = \frac{{ - 6}}{{{{\left( {x + 3} \right)}^2}}} + \frac{2}{{\sqrt {4x} }}\)
\(f'\left( 1 \right) = \frac{{ - 6}}{{{{\left( {1 + 3} \right)}^2}}} + \frac{2}{{\sqrt {4.1} }} = \frac{5}{8}\).
A. $k = 1.$
B. $k = \frac{9}{2}.$
C. $k = - 3.$
D. $k = 3.$
Chọn D
Ta có $f'(x) = {\left( {k.{x^{\frac{1}{3}}} + \sqrt x } \right)^\prime } = k.\frac{1}{3}.\frac{1}{{\sqrt[3]{{{x^2}}}}} + \frac{1}{{2\sqrt x }}$
$f'(1) = \frac{3}{2} \Leftrightarrow \frac{1}{3}k + \frac{1}{2} = \frac{3}{2} \Leftrightarrow \frac{1}{3}k = 1 \Leftrightarrow k = 3$
Ta có $f'(x) = {\left( {k.{x^{\frac{1}{3}}} + \sqrt x } \right)^\prime } = k.\frac{1}{3}.\frac{1}{{\sqrt[3]{{{x^2}}}}} + \frac{1}{{2\sqrt x }}$
$f'(1) = \frac{3}{2} \Leftrightarrow \frac{1}{3}k + \frac{1}{2} = \frac{3}{2} \Leftrightarrow \frac{1}{3}k = 1 \Leftrightarrow k = 3$
A. 0.
B. 1.
C. 2.
D. Không tồn tại.
Chọn D
Tập xác định của hàm số là: \(D = \left( {0; + \infty } \right)\).
\(x = 0 \notin D \Rightarrow \)không tồn tại đạo hàm tại x = 0.
Tập xác định của hàm số là: \(D = \left( {0; + \infty } \right)\).
\(x = 0 \notin D \Rightarrow \)không tồn tại đạo hàm tại x = 0.
A. 6.
B. 3.
C. - 2
D. - 6.
Chọn A
Có \(f(x) = 2{x^3} + 1\)\( \Rightarrow \)\(f'(x) = 6{x^2}\)\( \Rightarrow \)\(f'( - 1)\)\( = \)\(6.{( - 1)^2}\)\( = \)\(6.\)
Có \(f(x) = 2{x^3} + 1\)\( \Rightarrow \)\(f'(x) = 6{x^2}\)\( \Rightarrow \)\(f'( - 1)\)\( = \)\(6.{( - 1)^2}\)\( = \)\(6.\)
A. \(f'(2) = \frac{2}{{\sqrt 3 }}.\)
B. \(f'(2) = \frac{{ - 2}}{{\sqrt 3 }}.\)
C. \(f'(2) = \frac{{ - 2}}{{\sqrt { - 3} }}.\)
D. Không tồn tại.
Đáp án D
Ta có $f'\left( x \right) = {\left( {\sqrt {1 - {x^2}} } \right)^\prime } = \frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} = \frac{{ - x}}{{\sqrt {1 - {x^2}} }}$
Không tồn tại \(f'\left( 2 \right)\).
Ta có $f'\left( x \right) = {\left( {\sqrt {1 - {x^2}} } \right)^\prime } = \frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} = \frac{{ - x}}{{\sqrt {1 - {x^2}} }}$
Không tồn tại \(f'\left( 2 \right)\).
A. $\frac{1}{2}.$
B. $ - \frac{1}{2}.$
C. – 2.
D. Không tồn tại.
Đáp án D
Ta có $f'\left( x \right) = {\left( {\frac{{2x}}{{x - 1}}} \right)^\prime } = \frac{{2\left( {x - 1} \right) - 2x}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}$
Suy ra không tồn tại \(f'\left( 1 \right)\).
Ta có $f'\left( x \right) = {\left( {\frac{{2x}}{{x - 1}}} \right)^\prime } = \frac{{2\left( {x - 1} \right) - 2x}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}$
Suy ra không tồn tại \(f'\left( 1 \right)\).
A. 4.
B. 8.
C. -4.
D. 24.
Đáp án D
Ta có $f'\left( x \right) = 2\left( {3{x^2} - 1} \right){\left( {3{x^2} - 1} \right)^\prime } = 12x\left( {3{x^2} - 1} \right) \Rightarrow f'\left( 1 \right) = 24$
Ta có $f'\left( x \right) = 2\left( {3{x^2} - 1} \right){\left( {3{x^2} - 1} \right)^\prime } = 12x\left( {3{x^2} - 1} \right) \Rightarrow f'\left( 1 \right) = 24$
A. $\frac{1}{2}.$
B. $ - \frac{1}{2}.$
C. $\frac{1}{{\sqrt 2 }}.$
D. $ - \frac{1}{{\sqrt 2 }}.$
Đáp án B
$f'\left( x \right) = - \frac{1}{{{x^2}}} \Rightarrow f'\left( {\sqrt 2 } \right) = - \frac{1}{2}$
$f'\left( x \right) = - \frac{1}{{{x^2}}} \Rightarrow f'\left( {\sqrt 2 } \right) = - \frac{1}{2}$
A. 14.
B. 24.
C. 15.
D. 4.
Ta có suy ra
Chọn D
Chọn D
Kiểu 2: TÍNH ĐẠO HÀN BẰNG CÔNG THỨC
Câu 1. Đạo hàm của hàm số y = 10 là:
A. 10.
B. - 10.
C. 0.
D. 10x
Chọn C
Có y = 10\( \Rightarrow \)\(y' = 0.\)
Có y = 10\( \Rightarrow \)\(y' = 0.\)
A. \(f'(x) = - a.\)
B. \(f'(x) = - b.\)
C. \(f'(x) = a.\)
D. \(f'(x) = b.\)
Chọn C
Có \(f(x) = ax + b\)\( \Rightarrow \)\(f'(x) = a.\)
Có \(f(x) = ax + b\)\( \Rightarrow \)\(f'(x) = a.\)
A. \(y' = 4{x^3} - 6x + 3\)
B. \(y' = 4{x^4} - 6x + 2\)
C. \(y' = 4{x^3} - 3x + 2\)
D. \(y' = 4{x^3} - 6x + 2\)
Chọn D
Ta có: \(y' = 4{x^3} - 6x + 2\)
Ta có: \(y' = 4{x^3} - 6x + 2\)
A. \(y' = - 2{x^2} + 4x + 1\)
B. \(y' = - 3{x^2} + 4x + 1\)
C. \(y' = - \frac{1}{3}{x^2} + 4x + 1\)
D. \(y' = - {x^2} + 4x + 1\)
Chọn D
Ta có \(y' = - {x^2} + 4x + 1\)
Ta có \(y' = - {x^2} + 4x + 1\)
A. \(y' = 5{\left( {1 - {x^3}} \right)^4}\).
B. \(y' = - 15{x^2}{\left( {1 - {x^3}} \right)^5}\).
C. \(y' = - 3{\left( {1 - {x^3}} \right)^4}\).
D. \(y' = - 5{x^2}{\left( {1 - {x^3}} \right)^4}\).
Chọn B
Ta có : \(y' = 5{\left( {1 - {x^3}} \right)^4}{\left( {1 - {x^3}} \right)^\prime } = - 15{x^2}{\left( {1 - {x^3}} \right)^4}\).
Ta có : \(y' = 5{\left( {1 - {x^3}} \right)^4}{\left( {1 - {x^3}} \right)^\prime } = - 15{x^2}{\left( {1 - {x^3}} \right)^4}\).
A. $f'\left( x \right) = a$.
B. $f'\left( x \right) = - a$.
C. $f'\left( x \right) = b$.
D. $f'\left( x \right) = - b$.
Chọn A
Sử dụng các công thức đạo hàm: ${\left( c \right)^\prime } = 0$với $c = const$; $x' = 1$; ${\left( {k.u} \right)^\prime } = k.u'$với $k = const$.
${\left( {{x^n}} \right)^\prime } = n.{x^{n - 1}}$với $n$là số nguyên dương ;${\left( {u + v} \right)^\prime } = u' + v'$;
Ta có $f'\left( x \right) = {\left( {ax + b} \right)^\prime } = ax' + b' = a$.
Sử dụng các công thức đạo hàm: ${\left( c \right)^\prime } = 0$với $c = const$; $x' = 1$; ${\left( {k.u} \right)^\prime } = k.u'$với $k = const$.
${\left( {{x^n}} \right)^\prime } = n.{x^{n - 1}}$với $n$là số nguyên dương ;${\left( {u + v} \right)^\prime } = u' + v'$;
Ta có $f'\left( x \right) = {\left( {ax + b} \right)^\prime } = ax' + b' = a$.
A. $ - 4x - 3$.
B. $ - 4x + 3$.
C. $4x + 3$.
D. $4x - 3$.
Chọn B
Sử dụng các công thức đạo hàm: $x' = 1$; ${\left( {k.u} \right)^\prime } = k.u'$;${\left( {{x^n}} \right)^\prime } = n.{x^{n - 1}}$;${\left( {u + v} \right)^\prime } = u' + v'$.
$f'\left( x \right) = {\left( { - 2{x^2} + 3x} \right)^\prime } = - 2{\left( {{x^2}} \right)^\prime } + 3x' = - 4x + 3$.
Sử dụng các công thức đạo hàm: $x' = 1$; ${\left( {k.u} \right)^\prime } = k.u'$;${\left( {{x^n}} \right)^\prime } = n.{x^{n - 1}}$;${\left( {u + v} \right)^\prime } = u' + v'$.
$f'\left( x \right) = {\left( { - 2{x^2} + 3x} \right)^\prime } = - 2{\left( {{x^2}} \right)^\prime } + 3x' = - 4x + 3$.
A. $y' = 10{x^9} - 28{x^6} + 16{x^3}.$
B. $y' = 10{x^9} - 14{x^6} + 16{x^3}.$
C. $y' = 10{x^9} + 16{x^3}.$
D. $y' = 7{x^6} - 6{x^3} + 16x.$
Đáp án A
Ta có $y' = 2.\left( {{x^5} - 2{x^2}} \right){\left( {{x^5} - 2{x^2}} \right)^\prime } = 2\left( {{x^5} - 2{x^2}} \right)\left( {5{x^4} - 4x} \right) = 10{x^9} - 28{x^6} + 16{x^3}.$
Ta có $y' = 2.\left( {{x^5} - 2{x^2}} \right){\left( {{x^5} - 2{x^2}} \right)^\prime } = 2\left( {{x^5} - 2{x^2}} \right)\left( {5{x^4} - 4x} \right) = 10{x^9} - 28{x^6} + 16{x^3}.$
A. \(4{(7x - 5)^3}.\)
B. \( - 28{(7x - 5)^3}.\)
C. \(28{(7x - 5)^3}.\)
D. $A = y'' + y = - 3\sin x - 2\cos x + 3\sin x + 2{\rm{cos}}x = 0$
Đáp án C
Vì $y' = 4{\left( {7x - 5} \right)^3}{\left( {7x - 5} \right)^\prime } = 28{\left( {7x - 5} \right)^3}.$
Vì $y' = 4{\left( {7x - 5} \right)^3}{\left( {7x - 5} \right)^\prime } = 28{\left( {7x - 5} \right)^3}.$
A. 4x - 3.
B. - 4x + 3.
C. 4x + 3.
D. - 4x - 3.
Đáp án B
$f\left( x \right) = - 2{x^2} + 3x \Rightarrow f'\left( x \right) = - 4x + 3$
$f\left( x \right) = - 2{x^2} + 3x \Rightarrow f'\left( x \right) = - 4x + 3$
A. $y' = 2016{({x^3} - 2{x^2})^2}^{015}.$
B. $y' = 2016{({x^3} - 2{x^2})^{2015}}(3{x^2} - 4x).$
C. $y' = 2016({x^3} - 2{x^2})(3{x^2} - 4x).$
D. $y' = 2016({x^3} - 2{x^2})(3{x^2} - 2x).$
Chọn B
Đặt \(u = {x^3} - 2{x^2}\)thì\(y = {u^{2016}},\)\({y'_u} = 2016.{u^{2015}},\)\({u'_x} = 3{x^2} - 4x.\)
Theo công thức tính đạo hàm của hàm số hợp, ta có: ${y'_x} = {y'_u}.{u'_x}$.
Vậy:$y'$$ = $$2016.{({x^3} - 2{x^2})^2}^{015}.(3{x^2} - 4x).$
Đặt \(u = {x^3} - 2{x^2}\)thì\(y = {u^{2016}},\)\({y'_u} = 2016.{u^{2015}},\)\({u'_x} = 3{x^2} - 4x.\)
Theo công thức tính đạo hàm của hàm số hợp, ta có: ${y'_x} = {y'_u}.{u'_x}$.
Vậy:$y'$$ = $$2016.{({x^3} - 2{x^2})^2}^{015}.(3{x^2} - 4x).$
A. $6{x^5} - 20{x^4} + 16{x^3}$.
B. $6{x^5} + 16{x^3}$.
C. $6{x^5} - 20{x^4} + 4{x^3}$.
D. $6{x^5} - 20{x^4} - 16{x^3}$.
Chọn A
Cách 1: Áp dụng công thức ${\left( {{u^n}} \right)^\prime }$
Ta có \(y' = 2.\left( {{x^3} - 2{x^2}} \right).{\left( {{x^3} - 2{x^2}} \right)^\prime } = 2\left( {{x^3} - 2{x^2}} \right).\left( {3{x^2} - 4x} \right)\)
\( = 6{x^5} - 8{x^4} - 12{x^4} + 16{x^3} = 6{x^5} - 20{x^4} + 16{x^3}\)
Cách 2 : Khai triển hằng đẳng thức :
Ta có: $y = {\left( {{x^3} - 2{x^2}} \right)^2} = {x^6} - 4{x^5} + 4{x^4}$$ \Rightarrow y' = 6{x^5} - 20{x^4} + 16{x^3}$
Cách 1: Áp dụng công thức ${\left( {{u^n}} \right)^\prime }$
Ta có \(y' = 2.\left( {{x^3} - 2{x^2}} \right).{\left( {{x^3} - 2{x^2}} \right)^\prime } = 2\left( {{x^3} - 2{x^2}} \right).\left( {3{x^2} - 4x} \right)\)
\( = 6{x^5} - 8{x^4} - 12{x^4} + 16{x^3} = 6{x^5} - 20{x^4} + 16{x^3}\)
Cách 2 : Khai triển hằng đẳng thức :
Ta có: $y = {\left( {{x^3} - 2{x^2}} \right)^2} = {x^6} - 4{x^5} + 4{x^4}$$ \Rightarrow y' = 6{x^5} - 20{x^4} + 16{x^3}$
A. $y' = 3{x^5} + \frac{3}{{{x^2}}} + \frac{1}{{\sqrt x }}.$
B. $y' = 6{x^5} + \frac{3}{{{x^2}}} + \frac{1}{{2\sqrt x }}.$
C. $y' = 3{x^5} - \frac{3}{{{x^2}}} + \frac{1}{{\sqrt x }}.$
D. $y' = 6{x^5} - \frac{3}{{{x^2}}} + \frac{1}{{2\sqrt x }}.$
Chọn A
\(y' = 3{x^5} + \frac{3}{{{x^2}}} + \frac{1}{{\sqrt x }}\).
\(y' = 3{x^5} + \frac{3}{{{x^2}}} + \frac{1}{{\sqrt x }}\).
A. $2\left( {3{x^2} - 1} \right)$.
B. $6\left( {3{x^2} - 1} \right)$.
C. $6x\left( {3{x^2} - 1} \right)$.
D. $12x\left( {3{x^2} - 1} \right)$.
:
Chọn D
Ta có: $y = {\left( {3{x^2} - 1} \right)^2} \Rightarrow y' = 2\left( {3{x^2} - 1} \right){\left( {3{x^2} - 1} \right)^\prime } = 12x\left( {3{x^2} - 1} \right).$
Chọn D
Ta có: $y = {\left( {3{x^2} - 1} \right)^2} \Rightarrow y' = 2\left( {3{x^2} - 1} \right){\left( {3{x^2} - 1} \right)^\prime } = 12x\left( {3{x^2} - 1} \right).$
A. y’ = 4x
B. y' = 3x$^2$ - 6x + 2
C. y' = 2x$^2$ - 2x + 4
D. y' = 6x$^2$ - 2x - 4
Chọn D
$y = \left( {{x^2} - 2} \right)\left( {2x - 1} \right) \to y' = 2x\left( {2x - 1} \right) + 2\left( {{x^2} - 2} \right)$
$y = \left( {{x^2} - 2} \right)\left( {2x - 1} \right) \to y' = 2x\left( {2x - 1} \right) + 2\left( {{x^2} - 2} \right)$
A. \(y' = ({x^7} + x)(7{x^6} + 1)\)
B. \(y' = 2({x^7} + x)\)
C. \(y' = 2(7{x^6} + 1)\)
D. \(y' = 2({x^7} + x)(7{x^6} + 1)\)
Đáp án D
A. \(y' = - {x^3} + 4x\)
B. \(y' = - {x^3} - 4x\)
C. \(y' = 12{x^3} + 4x\)
D. \(y' = - 12{x^3} + 4x\)
Đáp án D
A. \(y' = {x^4} - 3{x^2} - 2\)
B. \(y' = 5{x^4} - 3{x^2} - 2\)
C. \(y' = 15{x^4} - 3{x^2}\)
D. \(y' = 15{x^4} - 3{x^2} - 2\)
Chọn D
Ta có: \(y' = 2x(3{x^3} + 2x) + ({x^2} - 1)(9{x^2} + 2) = 15{x^4} - 3{x^2} - 2\)
Ta có: \(y' = 2x(3{x^3} + 2x) + ({x^2} - 1)(9{x^2} + 2) = 15{x^4} - 3{x^2} - 2\)
A. \(y' = 40{x^2} - 3{x^2} - 6x\)
B. \(y' = 40{x^3} - 3{x^2} - 6x\)
C. \(y' = 40{x^3} + 3{x^2} - 6x\)
D. \(y' = 40{x^3} - 3{x^2} - x\)
Chọn B
\(y = 10{x^4} - {x^3} - 3{x^2} \Rightarrow y' = 40{x^3} - 3{x^2} - 6x\)
\(y = 10{x^4} - {x^3} - 3{x^2} \Rightarrow y' = 40{x^3} - 3{x^2} - 6x\)
A. \(y' = 3{({x^2} + 5x + 6)^3} + 2(x + 3){(x + 2)^3}\)
B. \(y' = 2{({x^2} + 5x + 6)^2} + 3(x + 3){(x + 2)^3}\)
C. \(y' = 3({x^2} + 5x + 6) + 2(x + 3)(x + 2)\)
D. \(y' = 3{({x^2} + 5x + 6)^2} + 2(x + 3){(x + 2)^3}\)
Chọn D
\(y' = 3{({x^2} + 5x + 6)^2} + 2(x + 3){(x + 2)^3}\)
\(y' = 3{({x^2} + 5x + 6)^2} + 2(x + 3){(x + 2)^3}\)
A. \(\left( {{x^7} + x} \right)\left( {7{x^6} + 1} \right)\)
B. \(2\left( {7{x^6} + 1} \right)\)
C. \(2\left( {{x^7} + x} \right)\left( {{x^6} + 1} \right)\)
D. \(2\left( {{x^7} + x} \right)\left( {7{x^6} + 1} \right)\)
Chọn D
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/} = \alpha .{u^{\alpha - 1}}.u'\) (với \(u = {x^7} + x\) )
\(y' = 2\left( {{x^7} + x} \right).{\left( {{x^7} + x} \right)^/} = 2\left( {{x^7} + x} \right)\left( {7{x^6} + 1} \right)\)
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/} = \alpha .{u^{\alpha - 1}}.u'\) (với \(u = {x^7} + x\) )
\(y' = 2\left( {{x^7} + x} \right).{\left( {{x^7} + x} \right)^/} = 2\left( {{x^7} + x} \right)\left( {7{x^6} + 1} \right)\)
A. \(2\left( {2{x^3} - {x^2} + 6x + 1} \right)\left( {6{x^2} - 6x + 6} \right).\)
B. \(2\left( {2{x^3} - 3{x^2} + x + 1} \right)\left( {{x^2} - 6x + 6} \right).\)
C. \(2\left( {2{x^3} - 3{x^2} + 6x + 1} \right)\left( {{x^2} - 6x + 6} \right).\)
D. \(2\left( {2{x^3} - 3{x^2} + 6x + 1} \right)\left( {6{x^2} - 6x + 6} \right).\)
Chọn D
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\) với \(u = 2{x^3} - 3{x^2} - 6x + 1\)
\(y' = 2\left( {2{x^3} - 3{x^2} + 6x + 1} \right){\left( {2{x^3} - 3{x^2} + 6x + 1} \right)^/} = 2\left( {2{x^3} - 3{x^2} + 6x + 1} \right)\left( {6{x^2} - 6x + 6} \right).\)
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\) với \(u = 2{x^3} - 3{x^2} - 6x + 1\)
\(y' = 2\left( {2{x^3} - 3{x^2} + 6x + 1} \right){\left( {2{x^3} - 3{x^2} + 6x + 1} \right)^/} = 2\left( {2{x^3} - 3{x^2} + 6x + 1} \right)\left( {6{x^2} - 6x + 6} \right).\)
A. \(12x{\left( {1 - 2{x^2}} \right)^2}.\)
B. \( - 12x{\left( {1 - 2{x^2}} \right)^2}.\)
C. \( - 24x{\left( {1 - 2{x^2}} \right)^2}.\)
D. \(24x{\left( {1 - 2{x^2}} \right)^2}.\)
Chọn B
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = 1 - 2{x^2}\)
\(y' = 3{\left( {1 - 2{x^2}} \right)^2}{\left( {1 - 2{x^2}} \right)^/} = 3{\left( {1 - 2{x^2}} \right)^2}\left( { - 4x} \right) = - 12x{\left( {1 - 2{x^2}} \right)^2}.\)
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = 1 - 2{x^2}\)
\(y' = 3{\left( {1 - 2{x^2}} \right)^2}{\left( {1 - 2{x^2}} \right)^/} = 3{\left( {1 - 2{x^2}} \right)^2}\left( { - 4x} \right) = - 12x{\left( {1 - 2{x^2}} \right)^2}.\)
A. \({\left( {x - {x^2}} \right)^{31}}.\left( {1 - 2x} \right)\)
B. \(32{\left( {x - {x^2}} \right)^{31}}\)
C. \(32{\left( {1 - {x^2}} \right)^{31}}\)
D. \(32{\left( {x - {x^2}} \right)^{31}}.\left( {1 - 2x} \right)\)
Chọn D
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = x - {x^2}\)
\(y' = 32{\left( {x - {x^2}} \right)^{31}}.{\left( {x - {x^2}} \right)^/} = 32{\left( {x - {x^2}} \right)^{31}}.\left( {1 - 2x} \right)\)
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = x - {x^2}\)
\(y' = 32{\left( {x - {x^2}} \right)^{31}}.{\left( {x - {x^2}} \right)^/} = 32{\left( {x - {x^2}} \right)^{31}}.\left( {1 - 2x} \right)\)
A. \(4{\left( {{x^2} + x + 1} \right)^3}.\)
B. \({\left( {{x^2} + x + 1} \right)^3}.\left( {2x + 1} \right)\)
C. \({\left( {{x^2} + x + 1} \right)^3}.\)
D. \(4{\left( {{x^2} + x + 1} \right)^3}.\left( {2x + 1} \right)\)
Chọn D
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = {x^2} + x + 1\)
\(y' = 4{\left( {{x^2} + x + 1} \right)^3}.{\left( {{x^2} + x + 1} \right)^/} = 4{\left( {{x^2} + x + 1} \right)^3}.\left( {2x + 1} \right)\)
Sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)với \(u = {x^2} + x + 1\)
\(y' = 4{\left( {{x^2} + x + 1} \right)^3}.{\left( {{x^2} + x + 1} \right)^/} = 4{\left( {{x^2} + x + 1} \right)^3}.\left( {2x + 1} \right)\)
A. \(y' = {\left( {{x^2} - x + 1} \right)^2}\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + 2\left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\)
B. \(y' = {\left( {{x^2} - x + 1} \right)^2}\left( {{x^2} + x + 1} \right)\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} - x + 1} \right)} \right]\)
C. \(y' = {\left( {{x^2} - x + 1} \right)^2}\left( {{x^2} + x + 1} \right)\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + 2\left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\)
D. \(y' = {\left( {{x^2} - x + 1} \right)^2}\left( {{x^2} + x + 1} \right)\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) - 2\left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\)
Chọn C
Đầu tiên sử dụng quy tắc nhân.
\(y' = {\left[ {{{\left( {{x^2} - x + 1} \right)}^3}} \right]^/}{\left( {{x^2} + x + 1} \right)^2} + {\left[ {{{\left( {{x^2} + x + 1} \right)}^2}} \right]^/}{\left( {{x^2} - x + 1} \right)^3}.\)
Sau đó sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)
\(y' = 3{\left( {{x^2} - x + 1} \right)^2}{\left( {{x^2} - x + 1} \right)^/}\left( {{x^2} + x + 1} \right) + 2\left( {{x^2} + x + 1} \right){\left( {{x^2} + x + 1} \right)^/}{\left( {{x^2} - x + 1} \right)^3}\)
\(y' = 3{\left( {{x^2} - x + 1} \right)^2}\left( {2x - 1} \right){\left( {{x^2} + x + 1} \right)^2} + 2\left( {{x^2} + x + 1} \right)\left( {2x + 1} \right){\left( {{x^2} - x + 1} \right)^3}\)
\(y' = {\left( {{x^2} - x + 1} \right)^2}\left( {{x^2} + x + 1} \right)\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + 2\left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\).
Đầu tiên sử dụng quy tắc nhân.
\(y' = {\left[ {{{\left( {{x^2} - x + 1} \right)}^3}} \right]^/}{\left( {{x^2} + x + 1} \right)^2} + {\left[ {{{\left( {{x^2} + x + 1} \right)}^2}} \right]^/}{\left( {{x^2} - x + 1} \right)^3}.\)
Sau đó sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\)
\(y' = 3{\left( {{x^2} - x + 1} \right)^2}{\left( {{x^2} - x + 1} \right)^/}\left( {{x^2} + x + 1} \right) + 2\left( {{x^2} + x + 1} \right){\left( {{x^2} + x + 1} \right)^/}{\left( {{x^2} - x + 1} \right)^3}\)
\(y' = 3{\left( {{x^2} - x + 1} \right)^2}\left( {2x - 1} \right){\left( {{x^2} + x + 1} \right)^2} + 2\left( {{x^2} + x + 1} \right)\left( {2x + 1} \right){\left( {{x^2} - x + 1} \right)^3}\)
\(y' = {\left( {{x^2} - x + 1} \right)^2}\left( {{x^2} + x + 1} \right)\left[ {3\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) + 2\left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\).
A. \(y' = \left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\)
B. \(y' = 4\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\)
C. \(y' = 2\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 - 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\)
D. \(y' = 2\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\)
Chọn C
\(y' = {\left( {1 + 2x} \right)^/}\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right){\left( {2 + 3{x^2}} \right)^/}\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right){\left( {3 - 4{x^3}} \right)^/}\)\(y' = 2\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\).
\(y' = {\left( {1 + 2x} \right)^/}\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right){\left( {2 + 3{x^2}} \right)^/}\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right){\left( {3 - 4{x^3}} \right)^/}\)\(y' = 2\left( {2 + 3{x^2}} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {6x} \right)\left( {3 - 4{x^3}} \right) + \left( {1 + 2x} \right)\left( {2 + 3{x^2}} \right)\left( { - 12{x^2}} \right)\).
A. \(\frac{a}{c}\)
B. \(\frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\)
C. \(\frac{{ad + bc}}{{{{\left( {cx + d} \right)}^2}}}\)
D. \(\frac{{ad - bc}}{{\left( {cx + d} \right)}}\)
Chọn B
Ta có \(y' = \frac{{ad - cb}}{{{{(cx + d)}^2}}} = \frac{{\left| \begin{array}{l}a{\rm{ }}b\\c{\rm{ }}d\end{array} \right|}}{{{{(cx + d)}^2}}}\)
Ta có \(y' = \frac{{ad - cb}}{{{{(cx + d)}^2}}} = \frac{{\left| \begin{array}{l}a{\rm{ }}b\\c{\rm{ }}d\end{array} \right|}}{{{{(cx + d)}^2}}}\)
A. \( - \frac{3}{{{{\left( {x + 2} \right)}^2}}}\)
B. \(\frac{3}{{\left( {x + 2} \right)}}\)
C. \(\frac{3}{{{{\left( {x + 2} \right)}^2}}}\)
D. \(\frac{2}{{{{\left( {x + 2} \right)}^2}}}\)
Chọn C
Ta có \(y' = \frac{{(2x + 1)'(x + 2) - (x + 2)'(2x + 1)}}{{{{(x + 2)}^2}}} = \frac{3}{{{{(x + 2)}^2}}}\)
Ta có \(y' = \frac{{(2x + 1)'(x + 2) - (x + 2)'(2x + 1)}}{{{{(x + 2)}^2}}} = \frac{3}{{{{(x + 2)}^2}}}\)
A. \(\frac{7}{{{{(2x - 1)}^2}}}\).
B. \(\frac{1}{{{{(2x - 1)}^2}}}\).
C. \( - \frac{{13}}{{{{(2x - 1)}^2}}}\).
D. \(\frac{{13}}{{{{(2x - 1)}^2}}}\).
Chọn C
Ta có \(y' = \frac{{{{\left( {3x + 5} \right)}^\prime }.\left( {2x - 1} \right) - \left( {3x + 5} \right){{\left( {2x - 1} \right)}^\prime }}}{{{{\left( {2x - 1} \right)}^2}}}\)
\( = \frac{{3\left( {2x - 1} \right) - 2\left( {3x + 5} \right)}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{{ - 13}}{{{{\left( {2x - 1} \right)}^2}}}\)
Có thể dùng công thức \({\left( {\frac{{ax + b}}{{cx + d}}} \right)^\prime } = \frac{{a.d - b.c}}{{{{\left( {cx + d} \right)}^2}}}\)
Ta có \(y' = \frac{{{{\left( {3x + 5} \right)}^\prime }.\left( {2x - 1} \right) - \left( {3x + 5} \right){{\left( {2x - 1} \right)}^\prime }}}{{{{\left( {2x - 1} \right)}^2}}}\)
\( = \frac{{3\left( {2x - 1} \right) - 2\left( {3x + 5} \right)}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{{ - 13}}{{{{\left( {2x - 1} \right)}^2}}}\)
Có thể dùng công thức \({\left( {\frac{{ax + b}}{{cx + d}}} \right)^\prime } = \frac{{a.d - b.c}}{{{{\left( {cx + d} \right)}^2}}}\)
A. \(y' = 2\).
B. \(y' = - \frac{1}{{{{\left( {x - 1} \right)}^2}}}\).
C. \(y' = - \frac{3}{{{{\left( {x - 1} \right)}^2}}}\).
D. \(y' = \frac{1}{{{{\left( {x - 1} \right)}^2}}}\).
Chọn C
Ta có : \(y' = \frac{{2\left( {x - 1} \right) - \left( {2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 3}}{{{{\left( {x - 1} \right)}^2}}}\).
Ta có : \(y' = \frac{{2\left( {x - 1} \right) - \left( {2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 3}}{{{{\left( {x - 1} \right)}^2}}}\).
A. \(\frac{{{x^2} - 2x}}{{{{\left( {x - 1} \right)}^2}}}\)
B. \(\frac{{{x^2} + 2x}}{{{{\left( {x - 1} \right)}^2}}}\)
C. \(\frac{{{x^2} + 2x}}{{{{\left( {x + 1} \right)}^2}}}\)
D. \(\frac{{ - 2x - 2}}{{{{\left( {x - 1} \right)}^2}}}\)
Chọn A
Ta có \(y' = \frac{{(2x - 1)(x - 1) - ({x^2} - x + 1)}}{{{{(x - 1)}^2}}} = \frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}\)
Ta có \(y' = \frac{{(2x - 1)(x - 1) - ({x^2} - x + 1)}}{{{{(x - 1)}^2}}} = \frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}\)
A. \( = \frac{{aa'{x^2} + 2ab'x + bb' - a'c}}{{(a'x + b')}}\)
B. \( = \frac{{aa'{x^2} + 2ab'x + bb' - a'c}}{{{{(a'x + b')}^2}}}\)
C. \( = \frac{{aa'{x^2} - 2ab'x + bb' - a'c}}{{{{(a'x + b')}^2}}}\)
D. \( = \frac{{aa'{x^2} + 2ab'x - bb' - a'c}}{{{{(a'x + b')}^2}}}\)
Chọn D
Ta có: \(y' = \frac{{(2ax + b)(a'x + b') - a'(a{x^2} + bx + c)}}{{{{(a'x + b')}^2}}}\)
\( = \frac{{aa'{x^2} + 2ab'x + bb' - a'c}}{{{{(a'x + b')}^2}}}\).
Ta có: \(y' = \frac{{(2ax + b)(a'x + b') - a'(a{x^2} + bx + c)}}{{{{(a'x + b')}^2}}}\)
\( = \frac{{aa'{x^2} + 2ab'x + bb' - a'c}}{{{{(a'x + b')}^2}}}\).
A. \(\frac{{2{x^2} + 6x + 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}\)
B. \(\frac{{2{x^2} - 6x + 2}}{{{{\left( {{x^2} - 1} \right)}^4}}}\)
C. \(\frac{{2{x^2} - 6x - 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}\)
D. \(\frac{{2{x^2} - 6x + 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}\)
Chọn D
Ta có \(y' = \frac{{(2x - 2)({x^2} - 1) - 2x({x^2} - 2x + 2)}}{{{{({x^2} - 1)}^2}}} = \frac{{2{x^2} - 6x + 2}}{{{{({x^2} - 1)}^2}}}\)
Ta có \(y' = \frac{{(2x - 2)({x^2} - 1) - 2x({x^2} - 2x + 2)}}{{{{({x^2} - 1)}^2}}} = \frac{{2{x^2} - 6x + 2}}{{{{({x^2} - 1)}^2}}}\)
A. \(y' = \frac{{ - {x^2} + 2x}}{{{{\left( {1 - x} \right)}^2}}}\).
B. \(y' = \frac{{{x^2} - 2x}}{{{{\left( {1 - x} \right)}^2}}}\).
C. \(y' = - 2\left( {x - 2} \right)\).
D. \(y' = \frac{{{x^2} + 2x}}{{{{\left( {1 - x} \right)}^2}}}\).
Chọn A
Ta có : \(y' = \frac{{2\left( {x - 2} \right)\left( {1 - x} \right) - {{\left( {x - 2} \right)}^2}\left( { - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\) \( = \frac{{ - {x^2} + 2x}}{{{{\left( {1 - x} \right)}^2}}}\).
Ta có : \(y' = \frac{{2\left( {x - 2} \right)\left( {1 - x} \right) - {{\left( {x - 2} \right)}^2}\left( { - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\) \( = \frac{{ - {x^2} + 2x}}{{{{\left( {1 - x} \right)}^2}}}\).
A. \( - 1 - \frac{3}{{{{(x - 2)}^2}}}\).
B. \(1 + \frac{3}{{{{(x - 2)}^2}}}\).
C. \( - 1 + \frac{3}{{{{(x - 2)}^2}}}\).
D. \(1 - \frac{3}{{{{(x - 2)}^2}}}\).
Đáp án
C.
Ta có \(y' = \frac{{{{\left( { - {x^2} + 2x - 3} \right)}^\prime }\left( {x - 2} \right) - \left( { - {x^2} + 2x - 3} \right){{\left( {x - 2} \right)}^\prime }}}{{{{\left( {x - 2} \right)}^2}}}\).
\( = \frac{{\left( { - 2x + 2} \right)\left( {x - 2} \right) - \left( { - {x^2} + 2x - 3} \right).1}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{ - {x^2} + 4x - 1}}{{{{\left( {x - 2} \right)}^2}}} = - 1 + \frac{3}{{{{\left( {x - 2} \right)}^2}}}\).
C.
Ta có \(y' = \frac{{{{\left( { - {x^2} + 2x - 3} \right)}^\prime }\left( {x - 2} \right) - \left( { - {x^2} + 2x - 3} \right){{\left( {x - 2} \right)}^\prime }}}{{{{\left( {x - 2} \right)}^2}}}\).
\( = \frac{{\left( { - 2x + 2} \right)\left( {x - 2} \right) - \left( { - {x^2} + 2x - 3} \right).1}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{ - {x^2} + 4x - 1}}{{{{\left( {x - 2} \right)}^2}}} = - 1 + \frac{3}{{{{\left( {x - 2} \right)}^2}}}\).
A. 1+ \(\frac{3}{{{{(x + 2)}^2}}}\).
B. \(\frac{{{x^2} + 6x + 7}}{{{{(x + 2)}^2}}}\).
C. \(\frac{{{x^2} + 4x + 5}}{{{{(x + 2)}^2}}}\).
D. \(\frac{{{x^2} + 8x + 1}}{{{{(x + 2)}^2}}}\).
Đáp án
A.
\({y^\prime } = \frac{{{{\left( {{x^2} + 2x - 3} \right)}^\prime }\left( {x + 2} \right) - {{\left( {x + 2} \right)}^\prime }\left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{\left( {2x + 2} \right)\left( {x + 2} \right) - \left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}}\)
\(\frac{{\left( {2x + 2} \right)\left( {x + 2} \right) - \left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{{x^2} + 4x + 7}}{{{{\left( {x + 2} \right)}^2}}} = 1 + \frac{3}{{{{\left( {x + 2} \right)}^2}}}\).
A.
\({y^\prime } = \frac{{{{\left( {{x^2} + 2x - 3} \right)}^\prime }\left( {x + 2} \right) - {{\left( {x + 2} \right)}^\prime }\left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{\left( {2x + 2} \right)\left( {x + 2} \right) - \left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}}\)
\(\frac{{\left( {2x + 2} \right)\left( {x + 2} \right) - \left( {{x^2} + 2x - 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{{x^2} + 4x + 7}}{{{{\left( {x + 2} \right)}^2}}} = 1 + \frac{3}{{{{\left( {x + 2} \right)}^2}}}\).
A. $y' = \frac{{2x - 2}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}}.$
B. $y' = \frac{{ - 2x + 2}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}}.$
C. \(y' = (2x - 2)({x^2} - 2x + 5).\)
D. \(y' = \frac{1}{{2x - 2}}.\)
Đáp án B
Vì \(y' = - \frac{{{{\left( {{x^2} - 2x + 5} \right)}^\prime }}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}} = \frac{{ - 2x + 2}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}}.\)
Vì \(y' = - \frac{{{{\left( {{x^2} - 2x + 5} \right)}^\prime }}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}} = \frac{{ - 2x + 2}}{{{{\left( {{x^2} - 2x + 5} \right)}^2}}}.\)
A. $\frac{{ - \left( {4x + 1} \right)}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}.$
B. $\frac{{ - \left( {4x - 1} \right)}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}.$
C. $\frac{{ - 1}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}.$
D. $\frac{{\left( {4x + 1} \right)}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}.$
Đáp án A
$y = \frac{1}{{2{x^2} + x + 1}} \Rightarrow y' = \frac{{ - {{\left( {2{x^2} + x + 1} \right)}^\prime }}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}} = \frac{{ - \left( {4x + 1} \right)}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}$
$y = \frac{1}{{2{x^2} + x + 1}} \Rightarrow y' = \frac{{ - {{\left( {2{x^2} + x + 1} \right)}^\prime }}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}} = \frac{{ - \left( {4x + 1} \right)}}{{{{\left( {2{x^2} + x + 1} \right)}^2}}}$
(I) $f'\left( x \right) = \frac{{{x^2} - 2x - 1}}{{{{\left( {x - 1} \right)}^2}}}\,\,\forall x \ne 1$ (II) \(f'\left( x \right) > 0\,\,\forall x \ne 1.\)
Hãy chọn câu đúng:
A. Chỉ (I) đúng.
B. Chỉ (II) đúng.
C. Cả hai đều sai.
D. Cả hai đều đúng.
Đáp án B
$f\left( x \right) = x + 1 - \frac{2}{{x - 1}} \Rightarrow f'\left( x \right) = 1 + \frac{2}{{{{\left( {x - 1} \right)}^2}}} = \frac{{{x^2} - 2x + 3}}{{{{\left( {x - 1} \right)}^2}}} > 0\,\forall x \ne 1$
$f\left( x \right) = x + 1 - \frac{2}{{x - 1}} \Rightarrow f'\left( x \right) = 1 + \frac{2}{{{{\left( {x - 1} \right)}^2}}} = \frac{{{x^2} - 2x + 3}}{{{{\left( {x - 1} \right)}^2}}} > 0\,\forall x \ne 1$
$(I):f'(x) = 1 - \frac{1}{{{{(x - 1)}^2}}},$$\forall x \ne 1.$ $(II):f'(x) = \frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}},$$\forall x \ne 1.$
Hãy chọn câu đúng:
A. Chỉ $(I)$đúng.
B. Chỉ $(II)$đúng.
C. Cả $(I);$$(II)$đều sai.
D. Cả $(I);$$(II)$đều đúng.
Chọn D
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}$ta có:
$\forall x \ne 1$, ta có: $f(x) = \frac{{{x^2} + x - 1}}{{x - 1}}$$ \Rightarrow $$f'(x) = \frac{{({x^2} + x - 1)'.(x - 1) - (x - 1)'.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}$
$ \Rightarrow $$f'(x)$$ = $$\frac{{(2x + 1).(x - 1) - 1.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}$$ = $$\frac{{2{x^2} - 2x + x - 1 - {x^2} - x + 1}}{{{{(x - 1)}^2}}}$$ = $$\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}$$ \Rightarrow $$(II)$đúng.
Mặt khác:$f'(x)$$ = $$\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}} = \frac{{{x^2} - 2x + 1 - 1}}{{{{(x - 1)}^2}}} = \frac{{{{(x - 1)}^2} - 1}}{{{{(x - 1)}^2}}} = 1 - \frac{1}{{{{(x - 1)}^2}}}$$ \Rightarrow $$(I)$đúng.
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}$ta có:
$\forall x \ne 1$, ta có: $f(x) = \frac{{{x^2} + x - 1}}{{x - 1}}$$ \Rightarrow $$f'(x) = \frac{{({x^2} + x - 1)'.(x - 1) - (x - 1)'.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}$
$ \Rightarrow $$f'(x)$$ = $$\frac{{(2x + 1).(x - 1) - 1.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}$$ = $$\frac{{2{x^2} - 2x + x - 1 - {x^2} - x + 1}}{{{{(x - 1)}^2}}}$$ = $$\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}$$ \Rightarrow $$(II)$đúng.
Mặt khác:$f'(x)$$ = $$\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}} = \frac{{{x^2} - 2x + 1 - 1}}{{{{(x - 1)}^2}}} = \frac{{{{(x - 1)}^2} - 1}}{{{{(x - 1)}^2}}} = 1 - \frac{1}{{{{(x - 1)}^2}}}$$ \Rightarrow $$(I)$đúng.
A. \(\frac{{ - 9{x^2} - 4x + 1}}{{{{(x + 1)}^2}}}.\)
B. \(\frac{{ - 3{x^2} - 6x + 1}}{{{{(x + 1)}^2}}}.\)
C. \(1 - 6{x^2}.\)
D. \(\frac{{1 - 6{x^2}}}{{{{(x + 1)}^2}}}.\)
Chọn B
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}.$Có : \(y = \frac{{x(1 - 3x)}}{{x + 1}}\)\( = \)\(\frac{{ - 3{x^2} + x}}{{x + 1}}\), nên:
\(y' = \frac{{( - 3{x^2} + x)'.(x + 1) - (x + 1)'.( - 3{x^2} + x)}}{{{{(x + 1)}^2}}}\)\( = \)\(\frac{{( - 6x + 1).(x + 1) - 1.( - 3{x^2} + x)}}{{{{(x + 1)}^2}}}\)
\( \Rightarrow \) y' \( = \)\(\frac{{ - 6{x^2} - 6x + x + 1 + 3{x^2} - x}}{{{{(x + 1)}^2}}}\)\( = \)\(\frac{{ - 3{x^2} - 6x + 1}}{{{{(x + 1)}^2}}}.\)
Chọn B
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}.$Có : \(y = \frac{{x(1 - 3x)}}{{x + 1}}\)\( = \)\(\frac{{ - 3{x^2} + x}}{{x + 1}}\), nên:
\(y' = \frac{{( - 3{x^2} + x)'.(x + 1) - (x + 1)'.( - 3{x^2} + x)}}{{{{(x + 1)}^2}}}\)\( = \)\(\frac{{( - 6x + 1).(x + 1) - 1.( - 3{x^2} + x)}}{{{{(x + 1)}^2}}}\)
\( \Rightarrow \) y' \( = \)\(\frac{{ - 6{x^2} - 6x + x + 1 + 3{x^2} - x}}{{{{(x + 1)}^2}}}\)\( = \)\(\frac{{ - 3{x^2} - 6x + 1}}{{{{(x + 1)}^2}}}.\)
Chọn B
A. \(\frac{{ - 3{x^2} - 13x - 10}}{{{{({x^2} + 3)}^2}}}.\)
B. \(\frac{{ - {x^2} + x + 3}}{{{{({x^2} + 3)}^2}}}.\)
C. \(\frac{{ - {x^2} + 2x + 3}}{{{{({x^2} + 3)}^2}}}.\)
D. \(\frac{{ - 7{x^2} - 13x - 10}}{{{{({x^2} + 3)}^2}}}.\)
Chọn C
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}.$Ta có:
\(y = \frac{{ - 2{x^2} + x - 7}}{{{x^2} + 3}}\)\( \Rightarrow \)\(y' = \frac{{( - 2{x^2} + x - 7)'.({x^2} + 3) - ({x^2} + 3)'.( - 2{x^2} + x - 7)}}{{{{({x^2} + 3)}^2}}}\)
\( \Rightarrow \)\(y' = \frac{{( - 4x + 1).({x^2} + 3) - 2x.( - 2{x^2} + x - 7)}}{{{{({x^2} + 3)}^2}}}\)\( = \)\(\frac{{ - 4{x^3} - 12x + {x^2} + 3 + 4{x^3} - 2{x^2} + 14x}}{{{{({x^2} + 3)}^2}}}\)
\( \Rightarrow \)\(y' = \frac{{ - {x^2} + 2x + 3}}{{{{({x^2} + 3)}^2}}}.\)
Áp dụng công thức ${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}.$Ta có:
\(y = \frac{{ - 2{x^2} + x - 7}}{{{x^2} + 3}}\)\( \Rightarrow \)\(y' = \frac{{( - 2{x^2} + x - 7)'.({x^2} + 3) - ({x^2} + 3)'.( - 2{x^2} + x - 7)}}{{{{({x^2} + 3)}^2}}}\)
\( \Rightarrow \)\(y' = \frac{{( - 4x + 1).({x^2} + 3) - 2x.( - 2{x^2} + x - 7)}}{{{{({x^2} + 3)}^2}}}\)\( = \)\(\frac{{ - 4{x^3} - 12x + {x^2} + 3 + 4{x^3} - 2{x^2} + 14x}}{{{{({x^2} + 3)}^2}}}\)
\( \Rightarrow \)\(y' = \frac{{ - {x^2} + 2x + 3}}{{{{({x^2} + 3)}^2}}}.\)
A. \(\frac{{2{x^2} + 10x + 9}}{{{{({x^2} + 3x + 3)}^2}}}\).
B. \(\frac{{ - 2{x^2} - 10x - 9}}{{{{({x^2} + 3x + 3)}^2}}}\).
C. \(\frac{{{x^2} - 2x - 9}}{{{{({x^2} + 3x + 3)}^2}}}\).
D. \(\frac{{ - 2{x^2} - 5x - 9}}{{{{({x^2} + 3x + 3)}^2}}}\).
Chọn B
Ta có \(y' = \frac{{{{\left( {2x + 5} \right)}^\prime }.\left( {{x^2} + 3x + 3} \right) - \left( {2x + 5} \right){{\left( {{x^2} + 3x + 3} \right)}^\prime }}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\)\( = \frac{{2\left( {{x^2} + 3x + 3} \right) - \left( {2x + 5} \right).\left( {2x + 3} \right)}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}} = \frac{{2{x^2} + 6x + 6 - 4{x^2} - 6x - 10x - 15}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\)
\( = \frac{{ - 2{x^2} - 10x - 9}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\).
Ta có \(y' = \frac{{{{\left( {2x + 5} \right)}^\prime }.\left( {{x^2} + 3x + 3} \right) - \left( {2x + 5} \right){{\left( {{x^2} + 3x + 3} \right)}^\prime }}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\)\( = \frac{{2\left( {{x^2} + 3x + 3} \right) - \left( {2x + 5} \right).\left( {2x + 3} \right)}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}} = \frac{{2{x^2} + 6x + 6 - 4{x^2} - 6x - 10x - 15}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\)
\( = \frac{{ - 2{x^2} - 10x - 9}}{{{{\left( {{x^2} + 3x + 3} \right)}^2}}}\).
A. \(\frac{{ - 2x - 2}}{{{{({x^2} - 2x + 5)}^2}}}.\)
B. \(\frac{{ - 4x + 4}}{{{{({x^2} - 2x + 5)}^2}}}.\)
C. \(\frac{{ - 2x + 2}}{{{{({x^2} - 2x + 5)}^2}}}.\)
D. \(\frac{{2x + 2}}{{{{({x^2} - 2x + 5)}^2}}}.\)
Chọn C
\(y' = \frac{{ - (2x - 2)}}{{{{({x^2} - 2x + 5)}^2}}} = \frac{{ - 2x + 2}}{{{{({x^2} - 2x + 5)}^2}}}.\)
\(y' = \frac{{ - (2x - 2)}}{{{{({x^2} - 2x + 5)}^2}}} = \frac{{ - 2x + 2}}{{{{({x^2} - 2x + 5)}^2}}}.\)
A. $\frac{{2{x^2} + 8x + 6}}{{{{(x - 2)}^2}}}$.
B. $\frac{{2{x^2} - 8x + 6}}{{x - 2}}.$
C. $\frac{{2{x^2} - 8x + 6}}{{{{(x - 2)}^2}}}$.
D. $\frac{{2{x^2} + 8x + 6}}{{x - 2}}$.
Chọn C
Ta có \(y' = 2 - \frac{2}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2{x^2} - 8x + 6}}{{{{(x - 2)}^2}}}.\)
Ta có \(y' = 2 - \frac{2}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2{x^2} - 8x + 6}}{{{{(x - 2)}^2}}}.\)
A. \(\frac{1}{{{{(x + 3)}^2}{{(x - 1)}^2}}}\).
B. \(\frac{1}{{2x + 2}}\).
C. \( - \frac{{2x + 2}}{{{{({x^2} + 2x - 3)}^2}}}\).
D. \(\frac{{ - 4}}{{{{\left( {{x^2} + 2x - 3} \right)}^2}}}\).
Chọn C
Ta có : $y = \frac{1}{{(x - 1)(x + 3)}} = \frac{1}{{{x^2} + 2x - 3}}$\( \Rightarrow y' = - \frac{{{{\left( {{x^2} + 2x - 3} \right)}^\prime }}}{{{{\left( {{x^2} + 2x - 3} \right)}^2}}} = - \frac{{2x + 2}}{{{{\left( {{x^2} + 2x - 3} \right)}^2}}}.\)
Ta có : $y = \frac{1}{{(x - 1)(x + 3)}} = \frac{1}{{{x^2} + 2x - 3}}$\( \Rightarrow y' = - \frac{{{{\left( {{x^2} + 2x - 3} \right)}^\prime }}}{{{{\left( {{x^2} + 2x - 3} \right)}^2}}} = - \frac{{2x + 2}}{{{{\left( {{x^2} + 2x - 3} \right)}^2}}}.\)
A. \(\frac{{ - 13{x^2} - 10x + 1}}{{{{({x^2} - 5x + 2)}^2}}}\).
B. \(\frac{{ - 13{x^2} + 5x + 11}}{{{{({x^2} - 5x + 2)}^2}}}\).
C. \(\frac{{ - 13{x^2} + 5x + 1}}{{{{({x^2} - 5x + 2)}^2}}}.\)
D. \(\frac{{ - 13{x^2} + 10x + 1}}{{{{({x^2} - 5x + 2)}^2}}}.\)
Chọn D
Ta có: \(y = \frac{{2{x^2} + 3x - 1}}{{{x^2} - 5x + 2}}.\)
$\begin{array}{l} y' = \frac{{{{\left( {2{x^3} + 3x - 1} \right)}^,}\left( {{x^2} - 5x + 2} \right) - \left( {2{x^3} + 3x - 1} \right){{\left( {{x^2} - 5x + 2} \right)}^,}}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\\ = \frac{{\left( {6{x^2} + 3} \right)\left( {{x^2} - 5x + 2} \right) - \left( {2{x^3} + 3x - 1} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}} = \frac{{ - 13{x^2} + 10x + 1}}{{{{({x^2} - 5x + 2)}^2}}}. \end{array}$
Ta có: \(y = \frac{{2{x^2} + 3x - 1}}{{{x^2} - 5x + 2}}.\)
$\begin{array}{l} y' = \frac{{{{\left( {2{x^3} + 3x - 1} \right)}^,}\left( {{x^2} - 5x + 2} \right) - \left( {2{x^3} + 3x - 1} \right){{\left( {{x^2} - 5x + 2} \right)}^,}}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\\ = \frac{{\left( {6{x^2} + 3} \right)\left( {{x^2} - 5x + 2} \right) - \left( {2{x^3} + 3x - 1} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}} = \frac{{ - 13{x^2} + 10x + 1}}{{{{({x^2} - 5x + 2)}^2}}}. \end{array}$
A. $y = {x^2} - \frac{1}{x}.$
B. $y = 2 - \frac{2}{{{x^3}}}.$
C. $y = {x^2} + \frac{1}{x}.$
D. $y = 2 - \frac{1}{x}.$
Đáp án A
Vì $y' = {\left( {{x^2} - \frac{1}{x}} \right)^\prime } = 2x + \frac{1}{{{x^2}}}.$
Vì $y' = {\left( {{x^2} - \frac{1}{x}} \right)^\prime } = 2x + \frac{1}{{{x^2}}}.$
A. \(\frac{{ - 3}}{{{x^4}}} + \frac{1}{{{x^3}}}.\)
B. \(\frac{{ - 3}}{{{x^4}}} + \frac{2}{{{x^3}}}.\)
C. \(\frac{{ - 3}}{{{x^4}}} - \frac{2}{{{x^3}}}.\)
D. \(\frac{3}{{{x^4}}} - \frac{1}{{{x^3}}}.\)
Đáp án A
Ta có \(y' = {\left( {\frac{1}{{{x^3}}} - \frac{1}{{{x^2}}}} \right)^\prime } = - \frac{{3{x^2}}}{{{x^6}}} + \frac{{2x}}{{{x^4}}} = - \frac{3}{{{x^4}}} + \frac{2}{{{x^3}}}\)
Ta có \(y' = {\left( {\frac{1}{{{x^3}}} - \frac{1}{{{x^2}}}} \right)^\prime } = - \frac{{3{x^2}}}{{{x^6}}} + \frac{{2x}}{{{x^4}}} = - \frac{3}{{{x^4}}} + \frac{2}{{{x^3}}}\)
A. $y = \frac{{{x^3} - 1}}{x}$
B. $y = \frac{{3({x^2} + x)}}{{{x^3}}}$
C. $y = \frac{{{x^3} + 5x - 1}}{x}$
D. $y = \frac{{2{x^2} + x - 1}}{x}$
Đáp án A
Kiểm tra đáp án A $y = \frac{{{x^3} - 1}}{x} = {x^2} - \frac{1}{x} \Rightarrow y' = 2x + \frac{1}{{{x^2}}}$đúng.
Kiểm tra đáp án A $y = \frac{{{x^3} - 1}}{x} = {x^2} - \frac{1}{x} \Rightarrow y' = 2x + \frac{1}{{{x^2}}}$đúng.
A. \(y' = \left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 - \frac{4}{{3{x^3}}}} \right)\)
B. \(y' = 2\left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 + \frac{4}{{3{x^3}}}} \right)\)
C. \(y' = \left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 + \frac{4}{{3{x^3}}}} \right)\)
D. \(y' = 2\left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 - \frac{4}{{3{x^3}}}} \right)\)
Chọn D
Ta có: \(y' = 2\left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 - \frac{4}{{3{x^3}}}} \right)\)
Ta có: \(y' = 2\left( {x + \frac{2}{{3{x^2}}}} \right)\left( {1 - \frac{4}{{3{x^3}}}} \right)\)
A. \(y' = 3\left( {4 + \frac{{10}}{{{x^3}}}} \right){\left( {4x + \frac{5}{{{x^2}}}} \right)^2}\)
B. \(y' = 3\left( {4 - \frac{{10}}{{{x^3}}}} \right){\left( {4x - \frac{5}{{{x^2}}}} \right)^2}\)
C. \(y' = {\left( {4x + \frac{5}{{{x^2}}}} \right)^2}\)
D. \(y' = 3\left( {4 - \frac{{10}}{{{x^3}}}} \right){\left( {4x + \frac{5}{{{x^2}}}} \right)^2}\)
Chọn D
\(y' = 3\left( {4 - \frac{{10}}{{{x^3}}}} \right){\left( {4x + \frac{5}{{{x^2}}}} \right)^2}\)
\(y' = 3\left( {4 - \frac{{10}}{{{x^3}}}} \right){\left( {4x + \frac{5}{{{x^2}}}} \right)^2}\)
A. $\frac{{3{x^2} + 2x}}{{2\sqrt {3{x^3} + 2{x^2} + 1} }}$
B. $\frac{{3{x^2} + 2x + 1}}{{2\sqrt {3{x^3} + 2{x^2} + 1} }}$
C. $\frac{{9{x^2} + 4x}}{{\sqrt {3{x^3} + 2{x^2} + 1} }}$
D. $\frac{{9{x^2} + 4x}}{{2\sqrt {3{x^3} + 2{x^2} + 1} }}$
Chọn D
Công thức ${\left( {\sqrt u } \right)^,} = \frac{1}{{2\sqrt u }}.u'$
Công thức ${\left( {\sqrt u } \right)^,} = \frac{1}{{2\sqrt u }}.u'$
A. \(y' = \frac{{3{x^2} - 6x}}{{\sqrt {{x^3} - 3{x^2} + 2} }}\)
B. \(y' = \frac{{3{x^2} + 6x}}{{2\sqrt {{x^3} - 3{x^2} + 2} }}\)
C. \(y' = \frac{{3{x^2} - 6x}}{{2\sqrt {{x^3} - 3{x^2} - 2} }}\)
D. \(y' = \frac{{3{x^2} - 6x}}{{2\sqrt {{x^3} - 3{x^2} + 2} }}\)
Chọn D
\(y' = \frac{{3{x^2} - 6x}}{{2\sqrt {{x^3} - 3{x^2} + 2} }}\)
\(y' = \frac{{3{x^2} - 6x}}{{2\sqrt {{x^3} - 3{x^2} + 2} }}\)
A. $\frac{{ - 4x}}{{2\sqrt {1 - 2{x^2}} }}$
B. $\frac{1}{{2\sqrt {1 - 2{x^2}} }}$
C. $\frac{{2x}}{{\sqrt {1 - 2{x^2}} }}$
D. $\frac{{ - 2x}}{{\sqrt {1 - 2{x^2}} }}$
Chọn D. $y = \sqrt {1 - 2{x^2}} \to y' = \frac{{\left( {1 - {x^2}} \right)'}}{{2\sqrt {1 - 2{x^2}} }} = \frac{{ - 2x}}{{\sqrt {1 - 2{x^2}} }}$
A. $\frac{{3\sqrt x }}{2}$.
B. $\frac{{\sqrt x }}{{2x}}$.
C. $\sqrt x + \frac{{\sqrt x }}{2}$.
D. $\frac{{\sqrt x }}{2}$.
.
Chọn A
Ta có: \(f\left( x \right) = x\sqrt x = {x^{\frac{3}{2}}} \Rightarrow f'\left( x \right) = \frac{3}{2}{x^{\frac{1}{2}}} = \frac{3}{2}\sqrt x .\)
Chọn A
Ta có: \(f\left( x \right) = x\sqrt x = {x^{\frac{3}{2}}} \Rightarrow f'\left( x \right) = \frac{3}{2}{x^{\frac{1}{2}}} = \frac{3}{2}\sqrt x .\)
A. \(\frac{7}{2}\sqrt {{x^5}} - \frac{5}{{2\sqrt x }}.\)
B. \(3{x^2} - \frac{1}{{2\sqrt x }}.\)
C. \(3{x^2} - \frac{5}{{2\sqrt x }}.\)
D. \(\frac{7}{2}\sqrt[5]{{{x^2}}} - \frac{5}{{2\sqrt x }}.\)
Chọn A
\(y' = {\left( {{x^3} - 5} \right)^\prime }\sqrt x + \left( {{x^3} - 5} \right){\left( {\sqrt x } \right)^\prime } = 3{x^2}.\sqrt x + \left( {{x^3} - 5} \right)\frac{1}{{2\sqrt x }} = \frac{{7{x^3} - 5}}{{2\sqrt x }} = \frac{7}{2}\sqrt {{x^5}} - \frac{5}{{2\sqrt x }}\).
\(y' = {\left( {{x^3} - 5} \right)^\prime }\sqrt x + \left( {{x^3} - 5} \right){\left( {\sqrt x } \right)^\prime } = 3{x^2}.\sqrt x + \left( {{x^3} - 5} \right)\frac{1}{{2\sqrt x }} = \frac{{7{x^3} - 5}}{{2\sqrt x }} = \frac{7}{2}\sqrt {{x^5}} - \frac{5}{{2\sqrt x }}\).
A. $\frac{{x - 6{x^2}}}{{\sqrt {{x^2} - 4{x^3}} }}.$
B. $\frac{1}{{2\sqrt {{x^2} - 4{x^3}} }}.$
C. $\frac{{x - 12{x^2}}}{{2\sqrt {{x^2} - 4{x^3}} }}.$
D. $\frac{{x - 6{x^2}}}{{2\sqrt {{x^2} - 4{x^3}} }}.$
Chọn A
\(y' = \frac{{2x - 12{x^2}}}{{2\sqrt {{x^2} - 4{x^3}} }} = \frac{{x - 6{x^2}}}{{\sqrt {{x^2} - 4{x^3}} }}\).
\(y' = \frac{{2x - 12{x^2}}}{{2\sqrt {{x^2} - 4{x^3}} }} = \frac{{x - 6{x^2}}}{{\sqrt {{x^2} - 4{x^3}} }}\).
A. $\frac{{3x - 1}}{{\sqrt {3{x^2} - 2x + 1} }}.$
B. $\frac{{6x - 2}}{{\sqrt {3{x^2} - 2x + 1} }}.$
C. $\frac{{3{x^2} - 1}}{{\sqrt {3{x^2} - 2x + 1} }}.$
D. $\frac{1}{{2\sqrt {3{x^2} - 2x + 1} }}.$
Chọn A
Áp dụng công thức ${\left( {\sqrt u } \right)^\prime } = \frac{{u'}}{{2\sqrt u }}$, ta được:
$y = \sqrt {3{x^2} - 2x + 1} $$ \Rightarrow $$y' = \frac{{(3{x^2} - 2x + 1)'}}{{2\sqrt {3{x^2} - 2x + 1} }}$$ = $$\frac{{6x - 2}}{{2\sqrt {3{x^2} - 2x + 1} }}$$ = $$\frac{{3x - 1}}{{\sqrt {3{x^2} - 2x + 1} }}.$
Áp dụng công thức ${\left( {\sqrt u } \right)^\prime } = \frac{{u'}}{{2\sqrt u }}$, ta được:
$y = \sqrt {3{x^2} - 2x + 1} $$ \Rightarrow $$y' = \frac{{(3{x^2} - 2x + 1)'}}{{2\sqrt {3{x^2} - 2x + 1} }}$$ = $$\frac{{6x - 2}}{{2\sqrt {3{x^2} - 2x + 1} }}$$ = $$\frac{{3x - 1}}{{\sqrt {3{x^2} - 2x + 1} }}.$
A. \(\frac{{4x + 5}}{{2\sqrt {2{x^2} + 5x - 4} }}.\)
B. \(\frac{{4x + 5}}{{\sqrt {2{x^2} + 5x - 4} }}.\)
C. \(\frac{{2x + 5}}{{2\sqrt {2{x^2} + 5x - 4} }}.\)
D. \(\frac{{2x + 5}}{{\sqrt {2{x^2} + 5x - 4} }}.\)
Chọn A
Áp dụng công thức ${\left( {\sqrt u } \right)^\prime } = \frac{{u'}}{{2\sqrt u }}$, ta được:
\(y = \sqrt {2{x^2} + 5x - 4} \)\( \Rightarrow \)\(y' = \frac{{(2{x^2} + 5x - 4)'}}{{2\sqrt {2{x^2} + 5x - 4} }}\)\( = \)\(\frac{{4x + 5}}{{2\sqrt {2{x^2} + 5x - 4} }}.\)
Áp dụng công thức ${\left( {\sqrt u } \right)^\prime } = \frac{{u'}}{{2\sqrt u }}$, ta được:
\(y = \sqrt {2{x^2} + 5x - 4} \)\( \Rightarrow \)\(y' = \frac{{(2{x^2} + 5x - 4)'}}{{2\sqrt {2{x^2} + 5x - 4} }}\)\( = \)\(\frac{{4x + 5}}{{2\sqrt {2{x^2} + 5x - 4} }}.\)
A. \(\frac{{2{x^2} + 1}}{{2\sqrt {{x^2} + 1} }}\)
B. \(\frac{{{x^2} + 1}}{{\sqrt {{x^2} + 1} }}\)
C. \(\frac{{4{x^2} + 1}}{{\sqrt {{x^2} + 1} }}\)
D. \(\frac{{2{x^2} + 1}}{{\sqrt {{x^2} + 1} }}\)
Chọn D
Ta có: \(y' = x'\sqrt {{x^2} + 1} + \left( {\sqrt {{x^2} + 1} } \right)'x = \sqrt {{x^2} + 1} + \frac{{({x^2} + 1)'}}{{2\sqrt {{x^2} + 1} }}.x\)
\( = \sqrt {{x^2} + 1} + \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }} = \frac{{2{x^2} + 1}}{{\sqrt {{x^2} + 1} }}\).
Ta có: \(y' = x'\sqrt {{x^2} + 1} + \left( {\sqrt {{x^2} + 1} } \right)'x = \sqrt {{x^2} + 1} + \frac{{({x^2} + 1)'}}{{2\sqrt {{x^2} + 1} }}.x\)
\( = \sqrt {{x^2} + 1} + \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }} = \frac{{2{x^2} + 1}}{{\sqrt {{x^2} + 1} }}\).
A. $y' = \frac{{2x - 2}}{{\sqrt {{x^2} - 2x} }}.$
B. $y' = \frac{{3{x^2} - 4x}}{{\sqrt {{x^2} - 2x} }}.$
C. $y' = \frac{{2{x^2} - 3x}}{{\sqrt {{x^2} - 2x} }}.$
D. $y' = \frac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 2x} }}.$
Đáp án C
$y = x.\sqrt {{x^2} - 2x} \Rightarrow y' = \sqrt {{x^2} - 2x} + x.\frac{{2x - 2}}{{2\sqrt {{x^2} - 2x} }} = \frac{{{x^2} - 2x + {x^2} - x}}{{\sqrt {{x^2} - 2x} }} = \frac{{2{x^2} - 3x}}{{\sqrt {{x^2} - 2x} }}$
$y = x.\sqrt {{x^2} - 2x} \Rightarrow y' = \sqrt {{x^2} - 2x} + x.\frac{{2x - 2}}{{2\sqrt {{x^2} - 2x} }} = \frac{{{x^2} - 2x + {x^2} - x}}{{\sqrt {{x^2} - 2x} }} = \frac{{2{x^2} - 3x}}{{\sqrt {{x^2} - 2x} }}$
A. $f'\left( x \right) = \frac{1}{2}\sqrt x $.
B. $f'\left( x \right) = \frac{3}{2}\sqrt x $.
C. $f'\left( x \right) = \frac{1}{2}\frac{{\sqrt x }}{x}$.
D. $f'\left( x \right) = x + \frac{{\sqrt x }}{2}$.
Chọn B
$\left( {u.v} \right)' = u'.v + u.v'$; $\left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }}$; $x' = 1$.
Ta có $f'\left( x \right) = \left( {x\sqrt x } \right)' = x'.\sqrt x + x.\left( {\sqrt x } \right)' = \sqrt x + \frac{x}{{2\sqrt x }} = \sqrt x + \frac{1}{2}\sqrt x = \frac{3}{2}\sqrt x $.
$\left( {u.v} \right)' = u'.v + u.v'$; $\left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }}$; $x' = 1$.
Ta có $f'\left( x \right) = \left( {x\sqrt x } \right)' = x'.\sqrt x + x.\left( {\sqrt x } \right)' = \sqrt x + \frac{x}{{2\sqrt x }} = \sqrt x + \frac{1}{2}\sqrt x = \frac{3}{2}\sqrt x $.
A. \(\frac{{4{x^2} - 5x + 3}}{{2\sqrt {{x^2} + x + 1} }}\)
B. \(\frac{{4{x^2} + 5x - 3}}{{2\sqrt {{x^2} + x + 1} }}\)
C. \(\frac{{4{x^2} + 5x + 3}}{{\sqrt {{x^2} + x + 1} }}\)
D. \(\frac{{4{x^2} + 5x + 3}}{{2\sqrt {{x^2} + x + 1} }}\)
Chọn D
Ta có \(y' = \sqrt {{x^2} + x + 1} + (x + 1)\frac{{2x + 1}}{{2\sqrt {{x^2} + x + 1} }} = \frac{{4{x^2} + 5x + 3}}{{2\sqrt {{x^2} + x + 1} }}\)
Ta có \(y' = \sqrt {{x^2} + x + 1} + (x + 1)\frac{{2x + 1}}{{2\sqrt {{x^2} + x + 1} }} = \frac{{4{x^2} + 5x + 3}}{{2\sqrt {{x^2} + x + 1} }}\)
A. \(y' = 2x + \sqrt {x + 1} - \frac{x}{{2\sqrt {x + 1} }}\)
B. \(y' = 2x - \sqrt {x + 1} + \frac{x}{{2\sqrt {x + 1} }}\)
C. \(y' = \frac{x}{{2\sqrt {x + 1} }}\)
D. \(y' = 2x + \sqrt {x + 1} + \frac{x}{{2\sqrt {x + 1} }}\)
Chọn D
\(y' = 2x + \sqrt {x + 1} + \frac{x}{{2\sqrt {x + 1} }}\)
\(y' = 2x + \sqrt {x + 1} + \frac{x}{{2\sqrt {x + 1} }}\)
A. \(y' = - \frac{{{a^2}}}{{\sqrt {{{({a^2} - {x^2})}^3}} }}\)
B. \(y' = \frac{{{a^2}}}{{\sqrt {{{({a^2} + {x^2})}^3}} }}\)
C. \(y' = \frac{{2{a^2}}}{{\sqrt {{{({a^2} - {x^2})}^3}} }}\)
D. \(y' = \frac{{{a^2}}}{{\sqrt {{{({a^2} - {x^2})}^3}} }}\)
Chọn D
\(y' = \frac{{\sqrt {{a^2} - {x^2}} + \frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}}}{{({a^2} - {x^2})}} = \frac{{{a^2}}}{{\sqrt {{{({a^2} - {x^2})}^3}} }}\)
\(y' = \frac{{\sqrt {{a^2} - {x^2}} + \frac{{{x^2}}}{{\sqrt {{a^2} - {x^2}} }}}}{{({a^2} - {x^2})}} = \frac{{{a^2}}}{{\sqrt {{{({a^2} - {x^2})}^3}} }}\)
A. \(y' = \frac{3}{2}\frac{1}{{{x^2}\sqrt x }}\)
B. \(y' = - \frac{1}{{{x^2}\sqrt x }}\)
C. \(y' = \frac{1}{{{x^2}\sqrt x }}\)
D. \(y' = - \frac{3}{2}\frac{1}{{{x^2}\sqrt x }}\)
Chọn D
$y' = - \frac{{(x\sqrt x )'}}{{{x^3}}} = - \frac{3}{2}\frac{1}{{{x^2}\sqrt x }}$
$y' = - \frac{{(x\sqrt x )'}}{{{x^3}}} = - \frac{3}{2}\frac{1}{{{x^2}\sqrt x }}$
A. \(y' = \frac{{1 - 3x}}{{\sqrt {{{(1 - x)}^3}} }}\)
B. \(y' = \frac{{1 - 3x}}{{3\sqrt {{{(1 - x)}^3}} }}\)
C. \(y' = - \frac{1}{3}\frac{{1 - 3x}}{{2\sqrt {{{(1 - x)}^3}} }}\)
D. \(y' = \frac{{1 - 3x}}{{2\sqrt {{{(1 - x)}^3}} }}\)
Chọn D
\(y' = \frac{{\sqrt {1 - x} - \frac{{1 + x}}{{2\sqrt {1 - x} }}}}{{1 - x}} = \frac{{1 - 3x}}{{2\sqrt {{{(1 - x)}^3}} }}\)
\(y' = \frac{{\sqrt {1 - x} - \frac{{1 + x}}{{2\sqrt {1 - x} }}}}{{1 - x}} = \frac{{1 - 3x}}{{2\sqrt {{{(1 - x)}^3}} }}\)
A. \(f'\left( x \right) = \frac{{ - 2\left( {1 - \sqrt x } \right)}}{{{{\left( {1 + \sqrt x } \right)}^3}}}\).
B. \(f'\left( x \right) = \frac{{ - 2\left( {1 - \sqrt x } \right)}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}}\).
C. \(f'\left( x \right) = \frac{{2\left( {1 - \sqrt x } \right)}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\).
D. \(f'\left( x \right) = \frac{{2\left( {1 - \sqrt x } \right)}}{{1 + \sqrt x }}\).
Chọn B
Ta có : \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right){\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^\prime }\) $ = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)\frac{{ - 2}}{{{{\left( {1 + \sqrt x } \right)}^2}}}{\left( {\sqrt x } \right)^\prime } = - \frac{2}{{\sqrt x }}\frac{{1 - \sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^3}}}$.
Ta có : \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right){\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^\prime }\) $ = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)\frac{{ - 2}}{{{{\left( {1 + \sqrt x } \right)}^2}}}{\left( {\sqrt x } \right)^\prime } = - \frac{2}{{\sqrt x }}\frac{{1 - \sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^3}}}$.
A. $f'\left( x \right) = x + \frac{1}{x} - 2$.
B. $f'\left( x \right) = x - \frac{1}{{{x^2}}}$.
C. $f'\left( x \right) = \sqrt x - \frac{1}{{\sqrt x }}$.
D. $f'\left( x \right) = 1 - \frac{1}{{{x^2}}}$.
Chọn D
Sử dụng công thức đạo hàm hợp: $\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'$và $\left( {\frac{1}{u}} \right)' = - \frac{{u'}}{{{u^2}}}$.
$\begin{array}{l}
f'\left( x \right) = {\left[ {{{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)}^2}} \right]^,} = 2.\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right).{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^,} = 2.\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)\\
= 2.\frac{1}{{2\sqrt x }}\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\left( {1 + \frac{1}{x}} \right) = \left( {1 - \frac{1}{x}} \right)\left( {1 + \frac{1}{x}} \right) = 1 - \frac{1}{{{x^2}}}
\end{array}$
Sử dụng công thức đạo hàm hợp: $\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'$và $\left( {\frac{1}{u}} \right)' = - \frac{{u'}}{{{u^2}}}$.
$\begin{array}{l}
f'\left( x \right) = {\left[ {{{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)}^2}} \right]^,} = 2.\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right).{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^,} = 2.\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)\\
= 2.\frac{1}{{2\sqrt x }}\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\left( {1 + \frac{1}{x}} \right) = \left( {1 - \frac{1}{x}} \right)\left( {1 + \frac{1}{x}} \right) = 1 - \frac{1}{{{x^2}}}
\end{array}$
A. \(f'\left( x \right) = \frac{3}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\).
B. \(f'\left( x \right) = \frac{3}{2}\left( {\sqrt x + \frac{1}{{\sqrt x }} + \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\).
C. \(f'\left( x \right) = \frac{3}{2}\left( { - \sqrt x + \frac{1}{{\sqrt x }} + \frac{1}{{x\sqrt x }} - \frac{1}{{{x^2}\sqrt x }}} \right)\).
D. \(f'\left( x \right) = x\sqrt x - 3\sqrt x + \frac{3}{{\sqrt x }} - \frac{1}{{x\sqrt x }}\).
Chọn A
Sử dụng công thức đạo hàm hợp: $\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'$và $\left( {\frac{1}{u}} \right)' = - \frac{{u'}}{{{u^2}}}$.
·Ta có: \(f'\left( x \right)\)$ = 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}.\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)$$ = 3.\frac{1}{{2\sqrt x }}\left( {x - 2 + \frac{1}{x}} \right).\left( {1 + \frac{1}{x}} \right)$
$ = \frac{3}{{2\sqrt x }}\left( {x - 1 - \frac{1}{x} + \frac{1}{{{x^2}}}} \right)$\( = \frac{3}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\).
Sử dụng công thức đạo hàm hợp: $\left( {{u^n}} \right)' = n.{u^{n - 1}}.u'$và $\left( {\frac{1}{u}} \right)' = - \frac{{u'}}{{{u^2}}}$.
·Ta có: \(f'\left( x \right)\)$ = 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}.\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)$$ = 3.\frac{1}{{2\sqrt x }}\left( {x - 2 + \frac{1}{x}} \right).\left( {1 + \frac{1}{x}} \right)$
$ = \frac{3}{{2\sqrt x }}\left( {x - 1 - \frac{1}{x} + \frac{1}{{{x^2}}}} \right)$\( = \frac{3}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\).
A. \(\frac{x}{{({x^2} + 1)\sqrt {{x^2} + 1} }}\).
B. \( - \frac{x}{{({x^2} + 1)\sqrt {{x^2} + 1} }}\).
C. \(\frac{x}{{2({x^2} + 1)\sqrt {{x^2} + 1} }}\).
D. \( - \frac{{x({x^2} + 1)}}{{\sqrt {{x^2} + 1} }}\).
Đáp án
B.
$y' = {\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)^\prime } = \frac{{ - {{\left( {\sqrt {{x^2} + 1} } \right)}^\prime }}}{{{x^2} + 1}} = \frac{{ - {{\left( {{x^2} + 1} \right)}^\prime }}}{{2\sqrt {{x^2} + 1} \left( {{x^2} + 1} \right)}} = \frac{{ - x}}{{\sqrt {{x^2} + 1} \left( {{x^2} + 1} \right)}}$.
B.
$y' = {\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)^\prime } = \frac{{ - {{\left( {\sqrt {{x^2} + 1} } \right)}^\prime }}}{{{x^2} + 1}} = \frac{{ - {{\left( {{x^2} + 1} \right)}^\prime }}}{{2\sqrt {{x^2} + 1} \left( {{x^2} + 1} \right)}} = \frac{{ - x}}{{\sqrt {{x^2} + 1} \left( {{x^2} + 1} \right)}}$.
(I) $f\left( x \right) = \frac{x}{{\sqrt {x - 1} }} \Rightarrow f'\left( x \right) = \frac{{x - 2}}{{2\left( {x - 1} \right)\sqrt {x - 1} }}$.
(II) $f\left( x \right) = \frac{1}{{2\sqrt {x - 1} }} - \frac{1}{{2\left( {x - 1} \right)\sqrt {x - 1} }} = \frac{{x - 2}}{{2\left( {x - 1} \right)\sqrt {x - 1} }}$.
Cách nào đúng?
A. Chỉ (I).
B. Chỉ (II)
C. Cả hai đều sai.
D. Cả hai đều đúng.
Đáp án
D.
$\sqrt {x - 1} + \frac{1}{{\sqrt {x - 1} }} = \frac{x}{{\sqrt {x - 1} }}$.
Lại có ${\left( {\frac{x}{{\sqrt {x - 1} }}} \right)^\prime } = \frac{{\sqrt {x - 1} - \frac{x}{{2\sqrt {x - 1} }}}}{{x - 1}} = \frac{{x - 2}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}$nên cả hai đều đúng.
D.
$\sqrt {x - 1} + \frac{1}{{\sqrt {x - 1} }} = \frac{x}{{\sqrt {x - 1} }}$.
Lại có ${\left( {\frac{x}{{\sqrt {x - 1} }}} \right)^\prime } = \frac{{\sqrt {x - 1} - \frac{x}{{2\sqrt {x - 1} }}}}{{x - 1}} = \frac{{x - 2}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}$nên cả hai đều đúng.
(I) $f'\left( x \right) = \frac{{ - 2x\left( {1 + 6{x^2}} \right)}}{{\sqrt {1 + 2{x^2}} }}$ (II) $f\left( x \right).f'\left( x \right) = 2x\left( {12{x^4} - 4{x^2} - 1} \right)$
Mệnh đề nào đúng?
A. Chỉ (II).
B. Chỉ (I).
C. Cả hai đều sai.
D. Cả hai đều đúng.
Đáp án D
Ta có
$\begin{array}{l}f'\left( x \right) = {\left( {1 - 2{x^2}} \right)^\prime }\sqrt {1 + 2{x^2}} + \left( {1 - 2{x^2}} \right){\left( {\sqrt {1 + 2{x^2}} } \right)^\prime } = - 4x\sqrt {1 + 2{x^2}} + \left( {1 - 2{x^2}} \right)\frac{{2x}}{{\sqrt {1 + 2{x^2}} }}\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{ - 4x\left( {1 + 2{x^2}} \right) + \left( {1 - 2{x^2}} \right).2x}}{{\sqrt {1 + 2{x^2}} }} = \frac{{ - 2x - 12{x^3}}}{{\sqrt {1 + 2{x^2}} }} = \frac{{ - 2x\left( {1 + 6{x^2}} \right)}}{{\sqrt {1 + 2{x^2}} }}\end{array}$
Suy ra
$\begin{array}{l}f\left( x \right).f'\left( x \right) = \left( {1 - 2{x^2}} \right)\sqrt {1 + 2{x^2}} .\frac{{ - 2x\left( {1 + 6{x^2}} \right)}}{{\sqrt {1 + 2{x^2}} }} = - 2x\left( {1 - 2{x^2}} \right)\left( {1 + 6{x^2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 2x\left( { - 12{x^4} + 4{x^2} + 1} \right) = 2x\left( {12{x^4} - 4{x^2} - 1} \right)\end{array}$
Ta có
$\begin{array}{l}f'\left( x \right) = {\left( {1 - 2{x^2}} \right)^\prime }\sqrt {1 + 2{x^2}} + \left( {1 - 2{x^2}} \right){\left( {\sqrt {1 + 2{x^2}} } \right)^\prime } = - 4x\sqrt {1 + 2{x^2}} + \left( {1 - 2{x^2}} \right)\frac{{2x}}{{\sqrt {1 + 2{x^2}} }}\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{ - 4x\left( {1 + 2{x^2}} \right) + \left( {1 - 2{x^2}} \right).2x}}{{\sqrt {1 + 2{x^2}} }} = \frac{{ - 2x - 12{x^3}}}{{\sqrt {1 + 2{x^2}} }} = \frac{{ - 2x\left( {1 + 6{x^2}} \right)}}{{\sqrt {1 + 2{x^2}} }}\end{array}$
Suy ra
$\begin{array}{l}f\left( x \right).f'\left( x \right) = \left( {1 - 2{x^2}} \right)\sqrt {1 + 2{x^2}} .\frac{{ - 2x\left( {1 + 6{x^2}} \right)}}{{\sqrt {1 + 2{x^2}} }} = - 2x\left( {1 - 2{x^2}} \right)\left( {1 + 6{x^2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 2x\left( { - 12{x^4} + 4{x^2} + 1} \right) = 2x\left( {12{x^4} - 4{x^2} - 1} \right)\end{array}$
A. \( - 14{x^6} + 2\sqrt x .\)
B. \( - 14{x^6} + \frac{2}{{\sqrt x }}.\)
C. \( - 14{x^6} + \frac{1}{{2\sqrt x }}.\)
D. \( - 14{x^6} + \frac{1}{{\sqrt x }}.\)
Đáp án C
Ta có \(y' = {\left( { - 2{x^7} + \sqrt x } \right)^\prime } = - 14{x^6} + \frac{1}{{2\sqrt x }}\)
Ta có \(y' = {\left( { - 2{x^7} + \sqrt x } \right)^\prime } = - 14{x^6} + \frac{1}{{2\sqrt x }}\)
A. $y' = \frac{5}{{{{\left( {2x - 1} \right)}^2}}}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
B. $y' = \frac{1}{2}.\frac{5}{{{{\left( {2x - 1} \right)}^2}}}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
C. $y' = \frac{1}{2}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
D. $y' = \frac{1}{2}.\frac{5}{{{{\left( {x + 2} \right)}^2}}}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
Đáp án
D.
Ta có $y' = \frac{1}{{2\sqrt {\frac{{2x - 1}}{{x + 2}}} }}.{\left( {\frac{{2x - 1}}{{x + 2}}} \right)^\prime } = \frac{1}{2}.\frac{5}{{{{\left( {x + 2} \right)}^2}}}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
D.
Ta có $y' = \frac{1}{{2\sqrt {\frac{{2x - 1}}{{x + 2}}} }}.{\left( {\frac{{2x - 1}}{{x + 2}}} \right)^\prime } = \frac{1}{2}.\frac{5}{{{{\left( {x + 2} \right)}^2}}}.\sqrt {\frac{{x + 2}}{{2x - 1}}} .$
A. \(\frac{1}{{2\sqrt x {{(1 - 2x)}^2}}}\).
B. \(\frac{1}{{ - 4\sqrt x }}\).
C. \(\frac{{1 - 2x}}{{2\sqrt x {{(1 - 2x)}^2}}}\).
D. \(\frac{{1 + 2x}}{{2\sqrt x {{(1 - 2x)}^2}}}\).
:
Chọn D
Ta có
\(y' = \frac{{{{\left( {\sqrt x } \right)}^\prime }.\left( {1 - 2x} \right) - {{\left( {1 - 2x} \right)}^\prime }.\sqrt x }}{{{{\left( {1 - 2x} \right)}^2}}} = \frac{{\frac{1}{{2\sqrt x }}.\left( {1 - 2x} \right) + 2\sqrt x }}{{{{\left( {1 - 2x} \right)}^2}}}\)
$ = \frac{{\frac{{1 - 2x + 4x}}{{2\sqrt x }}}}{{{{\left( {1 - 2x} \right)}^2}}} = \frac{{1 + 2x}}{{2\sqrt x {{\left( {1 - 2x} \right)}^2}}}$.
Chọn D
Ta có
\(y' = \frac{{{{\left( {\sqrt x } \right)}^\prime }.\left( {1 - 2x} \right) - {{\left( {1 - 2x} \right)}^\prime }.\sqrt x }}{{{{\left( {1 - 2x} \right)}^2}}} = \frac{{\frac{1}{{2\sqrt x }}.\left( {1 - 2x} \right) + 2\sqrt x }}{{{{\left( {1 - 2x} \right)}^2}}}\)
$ = \frac{{\frac{{1 - 2x + 4x}}{{2\sqrt x }}}}{{{{\left( {1 - 2x} \right)}^2}}} = \frac{{1 + 2x}}{{2\sqrt x {{\left( {1 - 2x} \right)}^2}}}$.
A. $y' = \frac{{13}}{{{{\left( {x + 5} \right)}^2}}} - \frac{1}{{\sqrt {2x} }}.$
B. $y' = \frac{{17}}{{{{\left( {x + 5} \right)}^2}}} - \frac{1}{{2\sqrt {2x} }}.$
C. $y' = \frac{{13}}{{{{\left( {x + 5} \right)}^2}}} - \frac{1}{{2\sqrt {2x} }}.$
D. $y' = \frac{{17}}{{{{\left( {x + 5} \right)}^2}}} - \frac{1}{{\sqrt {2x} }}.$
Chọn A
Cách 1:Ta có \(y' = \frac{{{{\left( {2x - 3} \right)}^\prime }.\left( {5 + x} \right) - \left( {2x - 3} \right).{{\left( {5 + x} \right)}^\prime }}}{{{{\left( {5 + x} \right)}^2}}} - \frac{{{{\left( {2x} \right)}^\prime }}}{{2\sqrt {2x} }}\)
\( = \frac{{2\left( {5 + x} \right) - \left( {2x - 3} \right)}}{{{{\left( {5 + x} \right)}^2}}} - \frac{2}{{2\sqrt {2x} }}.\)\( = \frac{{10 + 2x - 2x + 3}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }} = \frac{{13}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }}.\)
Cách 2: Ta có \(y' = \frac{{2.5 + 3.1}}{{{{\left( {5 + x} \right)}^2}}} - \frac{{{{\left( {2x} \right)}^\prime }}}{{2\sqrt {2x} }} = \frac{{13}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }}.\)
Có thể dùng công thức \({\left( {\frac{{ax + b}}{{cx + d}}} \right)^\prime } = \frac{{a.d - b.c}}{{{{\left( {cx + d} \right)}^2}}}\).
Cách 1:Ta có \(y' = \frac{{{{\left( {2x - 3} \right)}^\prime }.\left( {5 + x} \right) - \left( {2x - 3} \right).{{\left( {5 + x} \right)}^\prime }}}{{{{\left( {5 + x} \right)}^2}}} - \frac{{{{\left( {2x} \right)}^\prime }}}{{2\sqrt {2x} }}\)
\( = \frac{{2\left( {5 + x} \right) - \left( {2x - 3} \right)}}{{{{\left( {5 + x} \right)}^2}}} - \frac{2}{{2\sqrt {2x} }}.\)\( = \frac{{10 + 2x - 2x + 3}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }} = \frac{{13}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }}.\)
Cách 2: Ta có \(y' = \frac{{2.5 + 3.1}}{{{{\left( {5 + x} \right)}^2}}} - \frac{{{{\left( {2x} \right)}^\prime }}}{{2\sqrt {2x} }} = \frac{{13}}{{{{\left( {5 + x} \right)}^2}}} - \frac{x}{{\sqrt {2x} }}.\)
Có thể dùng công thức \({\left( {\frac{{ax + b}}{{cx + d}}} \right)^\prime } = \frac{{a.d - b.c}}{{{{\left( {cx + d} \right)}^2}}}\).
A. $y' = 2\sqrt {{x^2} + x} - \frac{{4{x^2} - 1}}{{2\sqrt {{x^2} + x} }}.$
B. $y' = 2\sqrt {{x^2} + x} + \frac{{4{x^2} - 1}}{{\sqrt {{x^2} + x} }}.$
C. $y' = 2\sqrt {{x^2} + x} + \frac{{4{x^2} - 1}}{{2\sqrt {{x^2} + x} }}.$
D. $y' = 2\sqrt {{x^2} + x} + \frac{{4{x^2} + 1}}{{2\sqrt {{x^2} + x} }}.$
Chọn C
Ta có \(y' = {\left( {2x - 1} \right)^\prime }.\sqrt {{x^2} + x} + \left( {2x - 1} \right).{\left( {\sqrt {{x^2} + x} } \right)^\prime } = 2.\sqrt {{x^2} + x} + \frac{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{2\sqrt {{x^2} + x} }}\)\( = 2\sqrt {{x^2} + x} + \frac{{4{x^2} - 1}}{{2\sqrt {{x^2} + x} }}\)
Ta có \(y' = {\left( {2x - 1} \right)^\prime }.\sqrt {{x^2} + x} + \left( {2x - 1} \right).{\left( {\sqrt {{x^2} + x} } \right)^\prime } = 2.\sqrt {{x^2} + x} + \frac{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{2\sqrt {{x^2} + x} }}\)\( = 2\sqrt {{x^2} + x} + \frac{{4{x^2} - 1}}{{2\sqrt {{x^2} + x} }}\)
A. \(\frac{{2x}}{{\sqrt {{x^2} + 1} }}.\)
B. \(\frac{{1 + x}}{{\sqrt {{{({x^2} + 1)}^3}} }}.\)
C. \(\frac{{2(x + 1)}}{{\sqrt {{{({x^2} + 1)}^3}} }}.\)
D. \(\frac{{{x^2} - x + 1}}{{\sqrt {{{({x^2} + 1)}^3}} }}.\)
Chọn B
\(y' = \frac{{{{\left( {x - 1} \right)}^\prime }.\sqrt {{x^2} + 1} - \left( {x - 1} \right){{\left( {\sqrt {{x^2} + 1} } \right)}^\prime }}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2}}} = \frac{{\sqrt {{x^2} + 1} - \left( {x - 1} \right)\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2}}} = \frac{{{x^2} + 1 - {x^2} + x}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^3}}} = \frac{{1 + x}}{{\sqrt {{{({x^2} + 1)}^3}} }}.\)
\(y' = \frac{{{{\left( {x - 1} \right)}^\prime }.\sqrt {{x^2} + 1} - \left( {x - 1} \right){{\left( {\sqrt {{x^2} + 1} } \right)}^\prime }}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2}}} = \frac{{\sqrt {{x^2} + 1} - \left( {x - 1} \right)\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2}}} = \frac{{{x^2} + 1 - {x^2} + x}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^3}}} = \frac{{1 + x}}{{\sqrt {{{({x^2} + 1)}^3}} }}.\)
A. $y' = - \frac{1}{{{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}^2}}}.$
B. $y' = \frac{1}{{2\sqrt {x + 1} + 2\sqrt {x - 1} }}.$
C. $y' = \frac{1}{{4\sqrt {x + 1} }} + \frac{1}{{4\sqrt {x - 1} }}.$
D. $y' = \frac{1}{{2\sqrt {x + 1} }} + \frac{1}{{2\sqrt {x - 1} }}.$
Chọn C
Ta có: $y = \frac{1}{{\sqrt {x + 1} - \sqrt {x - 1} }} = \frac{{\sqrt {x + 1} + \sqrt {x - 1} }}{2}$
\( \Rightarrow y' = \frac{1}{2}{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)^\prime } = \frac{1}{2}\left( {\frac{1}{{2\sqrt {x + 1} }} + \frac{1}{{2\sqrt {x - 1} }}} \right) = \frac{1}{{4\sqrt {x + 1} }} + \frac{1}{{4\sqrt {x - 1} }}.\)
Ta có: $y = \frac{1}{{\sqrt {x + 1} - \sqrt {x - 1} }} = \frac{{\sqrt {x + 1} + \sqrt {x - 1} }}{2}$
\( \Rightarrow y' = \frac{1}{2}{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)^\prime } = \frac{1}{2}\left( {\frac{1}{{2\sqrt {x + 1} }} + \frac{1}{{2\sqrt {x - 1} }}} \right) = \frac{1}{{4\sqrt {x + 1} }} + \frac{1}{{4\sqrt {x - 1} }}.\)
A. $\sqrt x - \frac{1}{{\sqrt x }}.$
B. $1 + \frac{1}{{{x^2}}}.$
C. $x + \frac{1}{x} - 2.$
D. $1 - \frac{1}{{{x^2}}}.$
Chọn D
$f\left( x \right) = {\rm{ }}{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} = x + \frac{1}{x} - 2 \to f'\left( x \right) = 1 - \frac{1}{{{x^2}}}$
$f\left( x \right) = {\rm{ }}{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} = x + \frac{1}{x} - 2 \to f'\left( x \right) = 1 - \frac{1}{{{x^2}}}$
A. \(\frac{1}{{\sqrt {{x^2} + 1} }} + \frac{x}{{\sqrt {1 - {x^2}} }}.\)
B. \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{1}{{\sqrt {1 - {x^2}} }}.\)
C. \(\frac{1}{{\sqrt {{x^2} + 1} }} + \frac{1}{{\sqrt {1 - {x^2}} }}.\)
D. \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{x}{{\sqrt {1 - {x^2}} }}.\)
Đáp án D
\(y' = {\left( {\sqrt {{x^2} + 1} } \right)^/} - {\left( {\sqrt {1 - {x^2}} } \right)^/} = \frac{{{{\left( {{x^2} + 1} \right)}^/}}}{{2\sqrt {{x^2} + 1} }} - \frac{{{{\left( {1 - {x^2}} \right)}^/}}}{{2\sqrt {1 - {x^2}} }} = \frac{x}{{\sqrt {{x^2} + 1} }} + \frac{x}{{\sqrt {1 - {x^2}} }}.\)
\(y' = {\left( {\sqrt {{x^2} + 1} } \right)^/} - {\left( {\sqrt {1 - {x^2}} } \right)^/} = \frac{{{{\left( {{x^2} + 1} \right)}^/}}}{{2\sqrt {{x^2} + 1} }} - \frac{{{{\left( {1 - {x^2}} \right)}^/}}}{{2\sqrt {1 - {x^2}} }} = \frac{x}{{\sqrt {{x^2} + 1} }} + \frac{x}{{\sqrt {1 - {x^2}} }}.\)
A. \(\frac{1}{{\sqrt {\frac{{{x^2} + 1}}{x}} }}\left( {1 - \frac{1}{{{x^2}}}} \right)\)
B. \(\frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}\)
C. \(\frac{3}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}\left( {1 - \frac{1}{{{x^2}}}} \right)\)
D. \(\frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}\left( {1 - \frac{1}{{{x^2}}}} \right)\)
Đáp án D
Sử dụng công thức \({\left( {\sqrt u } \right)^/}\) với \(u = \frac{{{x^2} + 1}}{x}\)
\(y' = \frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}.{\left( {\frac{{{x^2} + 1}}{x}} \right)^/} = \frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}\left( {1 - \frac{1}{{{x^2}}}} \right)\)
Sử dụng công thức \({\left( {\sqrt u } \right)^/}\) với \(u = \frac{{{x^2} + 1}}{x}\)
\(y' = \frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}.{\left( {\frac{{{x^2} + 1}}{x}} \right)^/} = \frac{1}{{2\sqrt {\frac{{{x^2} + 1}}{x}} }}\left( {1 - \frac{1}{{{x^2}}}} \right)\)
A. \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{1}{{{{\left( {1 + \sqrt x } \right)}^2}}}\)
B. \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)
C. \(y' = \left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)
D. \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{1}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)
Đáp án B
Đầu tiên sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\) với \(u = \frac{{1 - \sqrt x }}{{1 + \sqrt x }}\)
\(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).{\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^/}\)
Tính \({\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^/} = \frac{{{{\left( {1 - \sqrt x } \right)}^/}\left( {1 + \sqrt x } \right) - {{\left( {1 + \sqrt x } \right)}^/}\left( {1 - \sqrt x } \right)}}{{{{\left( {1 + \sqrt x } \right)}^2}}}\)
\( = \frac{{\frac{{ - 1}}{{2\sqrt x }}\left( {1 + \sqrt x } \right) - \frac{1}{{2\sqrt x }}\left( {1 - x} \right)}}{{{{\left( {1 + \sqrt x } \right)}^2}}} = \frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)
Vậy \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\).
Đầu tiên sử dụng công thức \({\left( {{u^\alpha }} \right)^/}\) với \(u = \frac{{1 - \sqrt x }}{{1 + \sqrt x }}\)
\(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).{\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^/}\)
Tính \({\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right)^/} = \frac{{{{\left( {1 - \sqrt x } \right)}^/}\left( {1 + \sqrt x } \right) - {{\left( {1 + \sqrt x } \right)}^/}\left( {1 - \sqrt x } \right)}}{{{{\left( {1 + \sqrt x } \right)}^2}}}\)
\( = \frac{{\frac{{ - 1}}{{2\sqrt x }}\left( {1 + \sqrt x } \right) - \frac{1}{{2\sqrt x }}\left( {1 - x} \right)}}{{{{\left( {1 + \sqrt x } \right)}^2}}} = \frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\)
Vậy \(y' = 2\left( {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \right).\frac{{ - 1}}{{\sqrt x {{\left( {1 + \sqrt x } \right)}^2}}}\).
A. \(\frac{1}{{\sqrt {x - 1} }} + \frac{{ - 1}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}.\)
B. \(\frac{1}{{2\sqrt {x - 1} }} + \frac{{ - 1}}{{2\sqrt {x - 1} }}.\)
C. \(\frac{1}{{\sqrt {x - 1} }} + \frac{{ - 1}}{{\sqrt {x - 1} \left( {x - 1} \right)}}.\)
D. \(\frac{1}{{2\sqrt {x - 1} }} + \frac{{ - 1}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}.\)
Đáp án D
\(y' = {\left( {\sqrt {x - 1} } \right)^/} + {\left( {\frac{1}{{\sqrt {x - 1} }}} \right)^/} = \frac{1}{{2\sqrt {x - 1} }} + \frac{{ - {{\left( {\sqrt {x - 1} } \right)}^/}}}{{{{\left( {\sqrt {x - 1} } \right)}^2}}} = \frac{1}{{2\sqrt {x - 1} }} + \frac{{ - 1}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}.\)
\(y' = {\left( {\sqrt {x - 1} } \right)^/} + {\left( {\frac{1}{{\sqrt {x - 1} }}} \right)^/} = \frac{1}{{2\sqrt {x - 1} }} + \frac{{ - {{\left( {\sqrt {x - 1} } \right)}^/}}}{{{{\left( {\sqrt {x - 1} } \right)}^2}}} = \frac{1}{{2\sqrt {x - 1} }} + \frac{{ - 1}}{{2\sqrt {x - 1} \left( {x - 1} \right)}}.\)
A. \(5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt x .x}}} \right)\)
B. \(5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{\sqrt x }} + \frac{1}{{\sqrt x .x}}} \right)\)
C. \({\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt x .x}}} \right)\)
D. \(5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt x .x}}} \right)\)
Đáp án D
Bước đầu tiên sử dụng \({\left( {{u^\alpha }} \right)^/}\)với \(u = \sqrt x - \frac{1}{{\sqrt x }}\)
\(y' = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}.{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^/} = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}.\left( {\frac{1}{{2\sqrt x }} + \frac{{{{\left( {\sqrt x } \right)}^/}}}{{{{\left( {\sqrt x } \right)}^2}}}} \right)\)
\( = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt x .x}}} \right)\)
Bước đầu tiên sử dụng \({\left( {{u^\alpha }} \right)^/}\)với \(u = \sqrt x - \frac{1}{{\sqrt x }}\)
\(y' = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}.{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^/} = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}.\left( {\frac{1}{{2\sqrt x }} + \frac{{{{\left( {\sqrt x } \right)}^/}}}{{{{\left( {\sqrt x } \right)}^2}}}} \right)\)
\( = 5{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^4}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt x .x}}} \right)\)
A. \(\frac{{ - x}}{{2\sqrt {1 - x} \left( {1 - x} \right)}}.\)
B. \(\frac{{3 - x}}{{\sqrt {1 - x} \left( {1 - x} \right)}}.\)
C. \(\frac{3}{{2\sqrt {1 - x} \left( {1 - x} \right)}}.\)
D. \(\frac{{3 - x}}{{2\sqrt {1 - x} \left( {1 - x} \right)}}.\)
Đáp án D
Sử dụng \({\left( {\frac{u}{v}} \right)^/}\) được: \(y' = \frac{{{{\left( {1 + x} \right)}^/}\sqrt {1 - x} - {{\left( {\sqrt {1 - x} } \right)}^/}\left( {1 + x} \right)}}{{{{\left( {\sqrt {1 - x} } \right)}^2}}}\) \( = \frac{{\sqrt {1 - x} - \frac{{{{\left( {1 - x} \right)}^/}}}{{2\sqrt {1 - x} }}.\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}\)\( = \frac{{2\left( {1 - x} \right) + \left( {1 + x} \right)}}{{2\sqrt {1 - x} .\left( {1 - x} \right)}} = \frac{{3 - x}}{{2\sqrt {1 - x} \left( {1 - x} \right)}}.\)
Sử dụng \({\left( {\frac{u}{v}} \right)^/}\) được: \(y' = \frac{{{{\left( {1 + x} \right)}^/}\sqrt {1 - x} - {{\left( {\sqrt {1 - x} } \right)}^/}\left( {1 + x} \right)}}{{{{\left( {\sqrt {1 - x} } \right)}^2}}}\) \( = \frac{{\sqrt {1 - x} - \frac{{{{\left( {1 - x} \right)}^/}}}{{2\sqrt {1 - x} }}.\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}\)\( = \frac{{2\left( {1 - x} \right) + \left( {1 + x} \right)}}{{2\sqrt {1 - x} .\left( {1 - x} \right)}} = \frac{{3 - x}}{{2\sqrt {1 - x} \left( {1 - x} \right)}}.\)
A. \(\frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{2\sqrt x }}} \right)} \right].\)
B. \(\frac{1}{{\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 + \frac{1}{{\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{\sqrt x }}} \right)} \right].\)
C. \(\frac{1}{{\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{2\sqrt x }}} \right)} \right].\)
D. \(\frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 - \frac{1}{{2\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{2\sqrt x }}} \right)} \right].\)
Đáp án A
Đầu tiên áp dụng \(\sqrt u \) với \(u = x + \sqrt {x + \sqrt x } \)
\(y' = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}{\left( {x + \sqrt {x + \sqrt x } } \right)^/} = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\left( {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.{{\left( {x + \sqrt x } \right)}^/}} \right)\)
\( = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{2\sqrt x }}} \right)} \right].\)
Đầu tiên áp dụng \(\sqrt u \) với \(u = x + \sqrt {x + \sqrt x } \)
\(y' = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}{\left( {x + \sqrt {x + \sqrt x } } \right)^/} = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\left( {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.{{\left( {x + \sqrt x } \right)}^/}} \right)\)
\( = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}.\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}.\left( {1 + \frac{1}{{2\sqrt x }}} \right)} \right].\)
A. \(\frac{{ - x}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\)
B. \(\frac{{x + 8}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\)
C. \(\frac{{ - x + 8}}{{\left( {{x^2} + 3} \right)\sqrt {{x^2} + 2} }}\)
D. \(\frac{{ - x + 8}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\)
Đáp án D
\(y' = \frac{{{{\left( {4x + 1} \right)}^/}\sqrt {{x^2} + 2} - {{\left( {\sqrt {{x^2} + 2} } \right)}^/}.\left( {4x + 1} \right)}}{{{{\left( {\sqrt {{x^2} + 2} } \right)}^2}}} = \frac{{4.\sqrt {{x^2} + 2} - \frac{{{{\left( {{x^2} + 2} \right)}^/}}}{{2\sqrt {{x^2} + 2} }}.\left( {4x + 1} \right)}}{{\left( {{x^2} + 2} \right)}}\)
\( = \frac{{4\sqrt {{x^2} + 2} - \frac{x}{{\sqrt {{x^2} + 2} }}\left( {4x + 1} \right)}}{{{x^2} + 2}} = \frac{{4\left( {{x^2} + 2} \right) - x\left( {4x + 1} \right)}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }} = \frac{{ - x + 8}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\)
\(y' = \frac{{{{\left( {4x + 1} \right)}^/}\sqrt {{x^2} + 2} - {{\left( {\sqrt {{x^2} + 2} } \right)}^/}.\left( {4x + 1} \right)}}{{{{\left( {\sqrt {{x^2} + 2} } \right)}^2}}} = \frac{{4.\sqrt {{x^2} + 2} - \frac{{{{\left( {{x^2} + 2} \right)}^/}}}{{2\sqrt {{x^2} + 2} }}.\left( {4x + 1} \right)}}{{\left( {{x^2} + 2} \right)}}\)
\( = \frac{{4\sqrt {{x^2} + 2} - \frac{x}{{\sqrt {{x^2} + 2} }}\left( {4x + 1} \right)}}{{{x^2} + 2}} = \frac{{4\left( {{x^2} + 2} \right) - x\left( {4x + 1} \right)}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }} = \frac{{ - x + 8}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\)
A. \(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
B. \(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{2{x^3} - {x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
C. \(y' = \frac{1}{{\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
D. \(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
Đáp án D
\(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^/}\)
Ta có: \({\left( {\frac{{{x^3}}}{{x - 1}}} \right)^/} = \frac{{{{\left( {{x^3}} \right)}^/}\left( {x - 1} \right) - {{\left( {x - 1} \right)}^/}.{x^3}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{3{x^2}\left( {x - 1} \right) - {x^3}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}\)
Vậy \(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
\(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^/}\)
Ta có: \({\left( {\frac{{{x^3}}}{{x - 1}}} \right)^/} = \frac{{{{\left( {{x^3}} \right)}^/}\left( {x - 1} \right) - {{\left( {x - 1} \right)}^/}.{x^3}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{3{x^2}\left( {x - 1} \right) - {x^3}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}\)
Vậy \(y' = \frac{1}{{2\sqrt {\frac{{{x^3}}}{{x - 1}}} }}.\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.\)
A. \(\frac{{\left( {x - 2} \right)}}{{2\sqrt {x - 2} }}.\)
B. \(\frac{{\left( {x - 2} \right)}}{{\sqrt {x - 2} }}.\)
C. \(\frac{{3\left( {x - 2} \right)}}{{\sqrt {x - 2} }}.\)
D. \(\frac{{3\left( {x - 2} \right)}}{{2\sqrt {x - 2} }}.\)
Đáp án D
Đầu tiên áp dụng \({\left( {\sqrt u } \right)^/}\) với \(u = {\left( {x - 2} \right)^3}\)
\(y' = \frac{1}{{2\sqrt {{{\left( {x - 2} \right)}^3}} }}.{\left( {{{\left( {x - 2} \right)}^3}} \right)^/} = \frac{1}{{2\sqrt {{{\left( {x - 2} \right)}^3}} }}.3.{\left( {x - 2} \right)^2} = \frac{{3\left( {x - 2} \right)}}{{2\sqrt {x - 2} }}.\)
Đầu tiên áp dụng \({\left( {\sqrt u } \right)^/}\) với \(u = {\left( {x - 2} \right)^3}\)
\(y' = \frac{1}{{2\sqrt {{{\left( {x - 2} \right)}^3}} }}.{\left( {{{\left( {x - 2} \right)}^3}} \right)^/} = \frac{1}{{2\sqrt {{{\left( {x - 2} \right)}^3}} }}.3.{\left( {x - 2} \right)^2} = \frac{{3\left( {x - 2} \right)}}{{2\sqrt {x - 2} }}.\)
A. \(\frac{{ - 6{{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{\sqrt {1 - 2x} }}.\)
B. \(\frac{{ - {{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{2\sqrt {1 - 2x} }}.\)
C. \(\frac{{ - {{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{\sqrt {1 - 2x} }}.\)
D. \(\frac{{ - 6{{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{2\sqrt {1 - 2x} }}.\)
Đáp án D
Bước đầu tiên áp dụng \({\left( {{u^\alpha }} \right)^/}\)với \(u = 1 + \sqrt {1 - 2x} \)
\(y' = 3{\left( {1 + \sqrt {1 - 2x} } \right)^2}.{\left( {1 + \sqrt {1 - 2x} } \right)^/} = 3{\left( {1 + \sqrt {1 - 2x} } \right)^2}.\frac{{{{\left( {1 - 2x} \right)}^/}}}{{2\sqrt {1 - 2x} }} = \frac{{ - 6{{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{2\sqrt {1 - 2x} }}.\)
Bước đầu tiên áp dụng \({\left( {{u^\alpha }} \right)^/}\)với \(u = 1 + \sqrt {1 - 2x} \)
\(y' = 3{\left( {1 + \sqrt {1 - 2x} } \right)^2}.{\left( {1 + \sqrt {1 - 2x} } \right)^/} = 3{\left( {1 + \sqrt {1 - 2x} } \right)^2}.\frac{{{{\left( {1 - 2x} \right)}^/}}}{{2\sqrt {1 - 2x} }} = \frac{{ - 6{{\left( {1 + \sqrt {1 - 2x} } \right)}^2}}}{{2\sqrt {1 - 2x} }}.\)
A. \(y' = \frac{{x + 2\sqrt {{x^2} + 1} }}{{\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\)
B. \(y' = \frac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\)
C. \(y' = \frac{{x + \sqrt {{x^2} + 1} }}{{2\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\)
D. \(y' = \frac{{x + 2\sqrt {{x^2} + 1} }}{{2\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\)
Chọn D
Ta có: \(y' = \frac{{\frac{x}{{\sqrt {{x^2} + 1} }} + 2}}{{2\sqrt {\sqrt {{x^2} + 1} + 2x - 1} }} = \frac{{x + 2\sqrt {{x^2} + 1} }}{{2\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\).
Ta có: \(y' = \frac{{\frac{x}{{\sqrt {{x^2} + 1} }} + 2}}{{2\sqrt {\sqrt {{x^2} + 1} + 2x - 1} }} = \frac{{x + 2\sqrt {{x^2} + 1} }}{{2\sqrt {({x^2} + 1)\left( {\sqrt {{x^2} + 1} + 2x - 1} \right)} }}\).
A. $f'\left( 1 \right) = 1$.
B. Hàm số có đạo hàm tại ${x_0} = 1$.
C. Hàm số liên tục tại ${x_0} = 1$.
D. $f'(x) = \left\{ \begin{array}{l}2x\,\,\,\,\,\,{\rm{khi}}\,\,\,x \ge 1\\2\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,\,x < 1\end{array} \right..$
:
Chọn A
Ta có: $f(1) = 1$
$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} {x^2} = 1$và $\mathop {\lim }\limits_{x \to {1^ - }} = \mathop {\lim }\limits_{x \to {1^ - }} (2x - 1) = 1$.
Vậy hàm số liên tục tại ${x_0} = 1$. C đúng.
Ta có: $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = 2$
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{(2x - 1) - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{2\left( {x - 1} \right)}}{{x - 1}} = 2$
Vậy hàm số có đạo hàm tại ${x_0} = 1$và $ \Rightarrow y' = - 2\sin 2x \Rightarrow y'' = - 4\cos 2x \Rightarrow y''\left( 0 \right) = - 4$
Chọn A
Ta có: $f(1) = 1$
$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} {x^2} = 1$và $\mathop {\lim }\limits_{x \to {1^ - }} = \mathop {\lim }\limits_{x \to {1^ - }} (2x - 1) = 1$.
Vậy hàm số liên tục tại ${x_0} = 1$. C đúng.
Ta có: $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = 2$
$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{(2x - 1) - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{2\left( {x - 1} \right)}}{{x - 1}} = 2$
Vậy hàm số có đạo hàm tại ${x_0} = 1$và $ \Rightarrow y' = - 2\sin 2x \Rightarrow y'' = - 4\cos 2x \Rightarrow y''\left( 0 \right) = - 4$
A. \(f'(x) = \left\{ \begin{array}{l}2x{\rm{ khi }}x < 1\\\frac{1}{{2\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\)
B. \(f'(x) = \left\{ \begin{array}{l}2x + 1{\rm{ khi }}x < 1\\ - \frac{1}{{\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\)
C. \(f'(x) = \left\{ \begin{array}{l}2x + 1{\rm{ khi }}x < 1\\\frac{1}{{\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\)
D. \(f'(x) = \left\{ \begin{array}{l}2x + 1{\rm{ khi }}x < 1\\\frac{1}{{2\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\)
:
Chọn D
Với \(x < 1\) ta có: \(f'(x) = 2x + 1\)
Với x > 1 ta có: \(f'(x) = \frac{1}{{2\sqrt {x - 1} }}\)
Tại x = 1 ta có:
\(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + x - 2}}{{x - 1}} = 3\)
\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {x - 1} }}{{x - 1}} = + \infty \) suy ra hàm số không có đạo
hàm tại x = 1
Vậy \(f'(x) = \left\{ \begin{array}{l}2x + 1{\rm{ khi }}x < 1\\\frac{1}{{2\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\).
Chọn D
Với \(x < 1\) ta có: \(f'(x) = 2x + 1\)
Với x > 1 ta có: \(f'(x) = \frac{1}{{2\sqrt {x - 1} }}\)
Tại x = 1 ta có:
\(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + x - 2}}{{x - 1}} = 3\)
\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {x - 1} }}{{x - 1}} = + \infty \) suy ra hàm số không có đạo
hàm tại x = 1
Vậy \(f'(x) = \left\{ \begin{array}{l}2x + 1{\rm{ khi }}x < 1\\\frac{1}{{2\sqrt {x - 1} }}{\rm{ khi }}x > 1\end{array} \right.\).
A. \(\left\{ \begin{array}{l}a = 13\\b = - 1\end{array} \right.\)
B. \(\left\{ \begin{array}{l}a = 3\\b = - 11\end{array} \right.\)
C. \(\left\{ \begin{array}{l}a = 23\\b = - 21\end{array} \right.\)
D. \(\left\{ \begin{array}{l}a = 3\\b = - 1\end{array} \right.\)
:
Chọn D
Với \(x \ne 1\) thì hàm số luôn có đạo hàm
Do đó hàm số có đạo hàm trên \(\mathbb{R}\)\( \Leftrightarrow \) hàm số có đạo hàm tại x = 1.
Ta có \(\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1;{\rm{ }}\mathop {\lim }\limits_{x \to {1^ + }} f(x) = a + b - 1\)
Hàm số liên tục trên \(\mathbb{R}\)\( \Leftrightarrow a + b - 1 = 1 \Leftrightarrow a + b = 2\)
Khi đó: \(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = 1;{\rm{ }}\)
\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{ - {x^2} + ax + 1 - a}}{{x - 1}} = a - 2\)
Nên hàm số có đạo hàm trên \(\mathbb{R}\) thì \(\left\{ \begin{array}{l}a + b = 2\\a - 2 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = - 1\end{array} \right.\).
Chọn D
Với \(x \ne 1\) thì hàm số luôn có đạo hàm
Do đó hàm số có đạo hàm trên \(\mathbb{R}\)\( \Leftrightarrow \) hàm số có đạo hàm tại x = 1.
Ta có \(\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1;{\rm{ }}\mathop {\lim }\limits_{x \to {1^ + }} f(x) = a + b - 1\)
Hàm số liên tục trên \(\mathbb{R}\)\( \Leftrightarrow a + b - 1 = 1 \Leftrightarrow a + b = 2\)
Khi đó: \(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}} = 1;{\rm{ }}\)
\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{ - {x^2} + ax + 1 - a}}{{x - 1}} = a - 2\)
Nên hàm số có đạo hàm trên \(\mathbb{R}\) thì \(\left\{ \begin{array}{l}a + b = 2\\a - 2 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = - 1\end{array} \right.\).
A. \(a = 0,b = 11\)
B. \(a = 10,b = 11\)
C. \(a = 20,b = 21\)
D. \(a = 0,b = 1\)
:
Chọn D
Tương tự như ý 1. ĐS: \(a = 0,b = 1\).
Chọn D
Tương tự như ý 1. ĐS: \(a = 0,b = 1\).
Kiểu 3: ĐẠO HÀM VÀ CÁC BÀI TOÁN GIẢI PT, BPT
Câu 1. Cho hàm số \(y = {x^3} - 3{x^2} - 9x - 5\). Phương trình \(y' = 0\) có nghiệm là:
A. \(\left\{ { - 1;2} \right\}\).
B. \(\left\{ { - 1;3} \right\}\).
C. \(\left\{ {0;4} \right\}\).
D. \(\left\{ {1;2} \right\}\).
Chọn B
Ta có : \(y' = 3{x^2} - 6x - 9\)
\(y' = 0 \Leftrightarrow 3{x^2} - 6x - 9 = 0 \Leftrightarrow x = - 1;x = 3\).
Ta có : \(y' = 3{x^2} - 6x - 9\)
\(y' = 0 \Leftrightarrow 3{x^2} - 6x - 9 = 0 \Leftrightarrow x = - 1;x = 3\).
A. k = 1
B. $k = - 3$.
C. $k = 3$.
D. $k = \frac{9}{2}$.
Chọn C
Ta có: $f\left( x \right) = k\sqrt[3]{x} + \sqrt x $$ \Rightarrow f'\left( x \right) = {\left( {k\sqrt[3]{x} + \sqrt x } \right)^\prime } = k{\left( {\sqrt[3]{x}} \right)^\prime } + {\left( {\sqrt x } \right)^\prime }$
Đặt \(y = \sqrt[3]{x} \Rightarrow {y^3} = x \Rightarrow 3{y^2}y' = 1 \Rightarrow y' = \frac{1}{{3{y^2}}} = \frac{1}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}}\).
$f'\left( x \right) = k{\left( {\sqrt[3]{x}} \right)^\prime } + {\left( {\sqrt x } \right)^\prime }$$ = \frac{k}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}} + \frac{1}{{2\sqrt x }}$.Vậy để $f'\left( 1 \right) = \frac{3}{2}$thì $\frac{k}{3} + \frac{1}{2} = \frac{3}{2} \Rightarrow k = 3$.
Ta có: $f\left( x \right) = k\sqrt[3]{x} + \sqrt x $$ \Rightarrow f'\left( x \right) = {\left( {k\sqrt[3]{x} + \sqrt x } \right)^\prime } = k{\left( {\sqrt[3]{x}} \right)^\prime } + {\left( {\sqrt x } \right)^\prime }$
Đặt \(y = \sqrt[3]{x} \Rightarrow {y^3} = x \Rightarrow 3{y^2}y' = 1 \Rightarrow y' = \frac{1}{{3{y^2}}} = \frac{1}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}}\).
$f'\left( x \right) = k{\left( {\sqrt[3]{x}} \right)^\prime } + {\left( {\sqrt x } \right)^\prime }$$ = \frac{k}{{3{{\left( {\sqrt[3]{x}} \right)}^2}}} + \frac{1}{{2\sqrt x }}$.Vậy để $f'\left( 1 \right) = \frac{3}{2}$thì $\frac{k}{3} + \frac{1}{2} = \frac{3}{2} \Rightarrow k = 3$.
A. $\left\{ { - 2\sqrt 2 } \right\}$.
B. $\left\{ {2;\sqrt 2 } \right\}$.
C. $\left\{ { - 4\sqrt 2 } \right\}$.
D. $\left\{ {2\sqrt 2 } \right\}$.
Chọn D
Ta có \(f'(x) = {x^2} - 4\sqrt 2 x + 8\)
\(f'(x) = 0 \Leftrightarrow {x^2} - 4\sqrt 2 x + 8 = 0 \Leftrightarrow x = 2\sqrt 2 \).
Ta có \(f'(x) = {x^2} - 4\sqrt 2 x + 8\)
\(f'(x) = 0 \Leftrightarrow {x^2} - 4\sqrt 2 x + 8 = 0 \Leftrightarrow x = 2\sqrt 2 \).
A. \(x = \frac{1}{8}.\)
B. \(x = \sqrt {\frac{1}{8}} .\)
C. \(x = \frac{1}{{64}}.\)
D. \(x = - \frac{1}{{64}}.\)
Chọn C
\(y' = 4 - \frac{1}{{2\sqrt x }}\)
\(y' = 0 \Leftrightarrow 4 - \frac{1}{{2\sqrt x }} = 0 \Leftrightarrow 8\sqrt x - 1 = 0 \Leftrightarrow \sqrt x = \frac{1}{8} \Rightarrow x = \frac{1}{{64}}\).
\(y' = 4 - \frac{1}{{2\sqrt x }}\)
\(y' = 0 \Leftrightarrow 4 - \frac{1}{{2\sqrt x }} = 0 \Leftrightarrow 8\sqrt x - 1 = 0 \Leftrightarrow \sqrt x = \frac{1}{8} \Rightarrow x = \frac{1}{{64}}\).
A. \(\left[ { - \sqrt 3 ;\sqrt 3 } \right].\)
B. \(\left[ { - \frac{1}{{\sqrt 3 }};\frac{1}{{\sqrt 3 }}} \right].\)
C. \(\left( { - \infty ; - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ; + \infty } \right).\)
D. \(\left( { - \infty ; - \frac{1}{{\sqrt 3 }}} \right] \cup \left[ {\frac{1}{{\sqrt 3 }}; + \infty } \right).\)
Chọn B
Ta có \(y = - 4{x^3} + 4x\)\( \Rightarrow y' = - 12{x^2} + 4\).
Nên \(y' \ge 0 \Leftrightarrow - 12{x^2} + 4 \ge 0 \Leftrightarrow x \in \left[ { - \frac{1}{{\sqrt 3 }};\frac{1}{{\sqrt 3 }}} \right].\)
Ta có \(y = - 4{x^3} + 4x\)\( \Rightarrow y' = - 12{x^2} + 4\).
Nên \(y' \ge 0 \Leftrightarrow - 12{x^2} + 4 \ge 0 \Leftrightarrow x \in \left[ { - \frac{1}{{\sqrt 3 }};\frac{1}{{\sqrt 3 }}} \right].\)
A. \(\left[ \begin{array}{l}x \le 0\\x \ge 1\end{array} \right.\)
B. \(x \le 1\)
C. \(x \ge 0\)
D. \(0 \le x \le 1\)
Chọn A
TXĐ: \(D = \mathbb{R}\)
Ta có: \(f'(x) = 6{x^2} - 6x\), suy ra \(f'(x) \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \le 0\\x \ge 1\end{array} \right.\)
TXĐ: \(D = \mathbb{R}\)
Ta có: \(f'(x) = 6{x^2} - 6x\), suy ra \(f'(x) \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \le 0\\x \ge 1\end{array} \right.\)
A. \(\left[ \begin{array}{l} - 1 < x < 0\\x > 1\end{array} \right.\)
B. - 1 < x < 0
C. x > 1
D. x < 0
Chọn A
TXĐ: \(D = \mathbb{R}\)
Ta có: \(f'(x) = - 8{x^3} + 8x\), suy ra \(f'(x) < 0 \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\x > 1\end{array} \right.\)
TXĐ: \(D = \mathbb{R}\)
Ta có: \(f'(x) = - 8{x^3} + 8x\), suy ra \(f'(x) < 0 \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\x > 1\end{array} \right.\)
A. \(x = \pm \frac{5}{3}\).
B. \(x = \pm \frac{3}{5}\).
C. x = 0.
D. \(x = \pm 5\).
:
Chọn A
Ta có: \(y' = - 9{x^2} + 25\)
\(y' = 0 \Leftrightarrow - 9{x^2} + 25 = 0 \Leftrightarrow x = \pm \frac{5}{3}.\)
Chọn A
Ta có: \(y' = - 9{x^2} + 25\)
\(y' = 0 \Leftrightarrow - 9{x^2} + 25 = 0 \Leftrightarrow x = \pm \frac{5}{3}.\)
A. $m \ge 1.$
B. $m \le - 1.$
C. $ - 1 \le m \le 1.$
D. $m \ge - 1.$
Chọn D
Có $f(x) = 2mx - m{x^3}$\( \Rightarrow \)\(f'(x) = 2m - 3m{x^2}.\)Nên$f'(1) \le 1$$ \Leftrightarrow $\(2m - 3m \le 1\)$ \Leftrightarrow $$m \ge - 1.$
Có $f(x) = 2mx - m{x^3}$\( \Rightarrow \)\(f'(x) = 2m - 3m{x^2}.\)Nên$f'(1) \le 1$$ \Leftrightarrow $\(2m - 3m \le 1\)$ \Leftrightarrow $$m \ge - 1.$
A. \(m \ge 3\)
B. \(m \ge 1\)
C. \(m \ge 4\)
D. \(m \ge 4\sqrt 2 \)
Chọn C
Ta có: \(y' = 3\left[ {(m - 1){x^2} - 2(m + 2)x - 2(m + 2)} \right]\)
Do đó \(y' \ge 0 \Leftrightarrow (m - 1){x^2} - 2(m + 2)x - 2(m + 2) \ge 0\) (1)
\( \bullet \) \(m = 1\) thì (1) \( \Leftrightarrow - 6x - 6 \ge 0 \Leftrightarrow x \le - 1\) nên \(m = 1\) (loại)
\( \bullet \) \(m \ne 1\) thì (1) đúng với \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a = m - 1 > 0\\\Delta ' \le 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}m > 1\\(m + 1)(4 - m) \le 0\end{array} \right. \Leftrightarrow m \ge 4\)
Vậy \(m \ge 4\) là những giá trị cần tìm.
Ta có: \(y' = 3\left[ {(m - 1){x^2} - 2(m + 2)x - 2(m + 2)} \right]\)
Do đó \(y' \ge 0 \Leftrightarrow (m - 1){x^2} - 2(m + 2)x - 2(m + 2) \ge 0\) (1)
\( \bullet \) \(m = 1\) thì (1) \( \Leftrightarrow - 6x - 6 \ge 0 \Leftrightarrow x \le - 1\) nên \(m = 1\) (loại)
\( \bullet \) \(m \ne 1\) thì (1) đúng với \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a = m - 1 > 0\\\Delta ' \le 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}m > 1\\(m + 1)(4 - m) \le 0\end{array} \right. \Leftrightarrow m \ge 4\)
Vậy \(m \ge 4\) là những giá trị cần tìm.
A. \(m \le \sqrt 2 \)
B. \(m \le 2\)
C. \(m \le 0\)
D. m < 0
Chọn C
Ta có: \(y' = m{x^2} - 2mx + 3m - 1\)
Nên \(y' \le 0 \Leftrightarrow m{x^2} - 2mx + 3m - 1 \le 0\) (2)
\( \bullet \) \(m = 0\) thì (1) trở thành: \( - 1 \le 0\) đúng với \(\forall x \in \mathbb{R}\)
\( \bullet \) \(m \ne 0\), khi đó (1) đúng với \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a = m < 0\\\Delta ' \le 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}m < 0\\m(1 - 2m) \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 0\\1 - 2m \ge 0\end{array} \right. \Leftrightarrow m < 0\)
Vậy \(m \le 0\) là những giá trị cần tìm.
Ta có: \(y' = m{x^2} - 2mx + 3m - 1\)
Nên \(y' \le 0 \Leftrightarrow m{x^2} - 2mx + 3m - 1 \le 0\) (2)
\( \bullet \) \(m = 0\) thì (1) trở thành: \( - 1 \le 0\) đúng với \(\forall x \in \mathbb{R}\)
\( \bullet \) \(m \ne 0\), khi đó (1) đúng với \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a = m < 0\\\Delta ' \le 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}m < 0\\m(1 - 2m) \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 0\\1 - 2m \ge 0\end{array} \right. \Leftrightarrow m < 0\)
Vậy \(m \le 0\) là những giá trị cần tìm.
A. $0 < x < 2$.
B. x < 1.
C. $x < 0$hoặc $x > 1.$
D. $x < 0$hoặc $x > 2.$
Chọn A
Ta có: \(f'\left( x \right) = 3{x^2} - 6x.\)
\(f'\left( x \right) < 0 \Leftrightarrow 3{x^2} - 6x < 0 \Leftrightarrow 0 < x < 2.\)
Ta có: \(f'\left( x \right) = 3{x^2} - 6x.\)
\(f'\left( x \right) < 0 \Leftrightarrow 3{x^2} - 6x < 0 \Leftrightarrow 0 < x < 2.\)
A. (- ∞; +∞)
B. (- ∞; 1/9)
C. ( 1/9; +∞)
D. Ø.
Chọn C
$\begin{array}{l} y = - 2\sqrt x + 3x \Rightarrow y' = 3 - \frac{1}{{\sqrt x }}\\ y' > 0 \Leftrightarrow 3 - \frac{1}{{\sqrt x }} > 0 \Leftrightarrow x > \frac{1}{9} \end{array}$
$\begin{array}{l} y = - 2\sqrt x + 3x \Rightarrow y' = 3 - \frac{1}{{\sqrt x }}\\ y' > 0 \Leftrightarrow 3 - \frac{1}{{\sqrt x }} > 0 \Leftrightarrow x > \frac{1}{9} \end{array}$
A. Ø.
B. (- ∞; 0]
C. [0; + ∞).
D. R
Chọn C
$y = {\left( {2{x^2} + 1} \right)^3} \Rightarrow y' = 12x{\left( {2{x^2} + 1} \right)^2} \Rightarrow y' \ge 0 \Leftrightarrow x \ge 0$
$y = {\left( {2{x^2} + 1} \right)^3} \Rightarrow y' = 12x{\left( {2{x^2} + 1} \right)^2} \Rightarrow y' \ge 0 \Leftrightarrow x \ge 0$
A. Ø.
B. (- ∞; 0).
C. ( 0; ∞).
D. (- ∞; 0]
Chọn D
$y = \sqrt {4{x^2} + 1} \Rightarrow y' = \frac{{4x}}{{\sqrt {4{x^2} + 1} }} \Rightarrow y' \le 0 \Leftrightarrow x \le 0$
$y = \sqrt {4{x^2} + 1} \Rightarrow y' = \frac{{4x}}{{\sqrt {4{x^2} + 1} }} \Rightarrow y' \le 0 \Leftrightarrow x \le 0$
A. 1.
B. 3.
C. \(\emptyset \).
D. \(\mathbb{R}\).
Chọn C
Tập xác định \(D = R\backslash \left\{ 1 \right\}\).
\({y^\prime } = \frac{3}{{{{\left( {1 - x} \right)}^2}}} > 0\forall x \in D\).
Tập xác định \(D = R\backslash \left\{ 1 \right\}\).
\({y^\prime } = \frac{3}{{{{\left( {1 - x} \right)}^2}}} > 0\forall x \in D\).
A. \(\mathbb{R}\backslash \left\{ 1 \right\}.\)
B. \(\emptyset .\)
C. \(\left( {1; + \infty } \right)\).
D. \(\mathbb{R}.\)
Đáp án A
\(\begin{array}{l}f'(x) = {\left( {\frac{{1 - 3x + {x^2}}}{{x - 1}}} \right)^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1 - 3x + {x^2}} \right)}^\prime }\left( {x - 1} \right) - \left( {1 - 3x + {x^2}} \right){{\left( {x - 1} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left( { - 3 + 2x} \right)\left( {x - 1} \right) - \left( {1 - 3x + {x^2}} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{{x^2} - 2x + 2}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {x - 1} \right)}^2} + 1}}{{{{\left( {x - 1} \right)}^2}}} > 0,\,\forall x \ne 1\end{array}\)
\(\begin{array}{l}f'(x) = {\left( {\frac{{1 - 3x + {x^2}}}{{x - 1}}} \right)^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1 - 3x + {x^2}} \right)}^\prime }\left( {x - 1} \right) - \left( {1 - 3x + {x^2}} \right){{\left( {x - 1} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left( { - 3 + 2x} \right)\left( {x - 1} \right) - \left( {1 - 3x + {x^2}} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{{x^2} - 2x + 2}}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {x - 1} \right)}^2} + 1}}{{{{\left( {x - 1} \right)}^2}}} > 0,\,\forall x \ne 1\end{array}\)
A. \(\left[ { - \frac{2}{9};0} \right].\)
B. \(\left[ { - \frac{9}{2};0} \right].\)
C. \(\left( { - \infty ; - \frac{9}{2}} \right] \cup \left[ {0; + \infty } \right).\)
D. \(\left( { - \infty ; - \frac{2}{9}} \right] \cup \left[ {0; + \infty } \right).\)
Đáp án A
\(\begin{array}{l}y = 3{x^3} + {x^2} + 1 \Rightarrow y' = 9{x^2} + 2x\\y' \le 0 \Rightarrow - \frac{2}{9} \le x \le 0\end{array}\)
\(\begin{array}{l}y = 3{x^3} + {x^2} + 1 \Rightarrow y' = 9{x^2} + 2x\\y' \le 0 \Rightarrow - \frac{2}{9} \le x \le 0\end{array}\)
A. Ø.
B. R\{0}.
C. ( - ∞; 0)
D. (0; +∞)
Lưu ý: Công thức đạo hàm nhanh $\left( {\frac{{ax + b}}{{cx + d}}} \right)' = \frac{{ab - bc}}{{{{\left( {cx + d} \right)}^2}}}$
$f'\left( x \right) < 0 \Leftrightarrow \frac{2}{{{{\left( {2x} \right)}^2}}} < 0$ vô nghiệm.
Chọn A
$f'\left( x \right) < 0 \Leftrightarrow \frac{2}{{{{\left( {2x} \right)}^2}}} < 0$ vô nghiệm.
Chọn A
A. \(x \ge \frac{1}{{\sqrt 3 }}\)
B. \(x > \frac{1}{{\sqrt 3 }}\)
C. \(x < \frac{1}{{\sqrt 3 }}\)
D. \(x \ge \frac{2}{{\sqrt 3 }}\)
TXĐ: \(D = \mathbb{R}\)
Ta có: \(f'(x) = 1 + \frac{x}{{\sqrt {{x^2} + 1} }} = \frac{{f(x)}}{{\sqrt {{x^2} + 1} }}\)
Mặt khác: \(f(x) > x + \sqrt {{x^2}} = x + \left| x \right| \ge 0,{\rm{ }}\forall x \in \mathbb{R}\)
Nên \(2xf'(x) - f(x) \ge 0 \Leftrightarrow \frac{{2xf(x)}}{{\sqrt {{x^2} + 1} }} - f(x) \ge 0\)
\( \Leftrightarrow 2x \ge \sqrt {{x^2} + 1} \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\3{x^2} \ge 1\end{array} \right. \Leftrightarrow x \ge \frac{1}{{\sqrt 3 }}\).
Ta có: \(f'(x) = 1 + \frac{x}{{\sqrt {{x^2} + 1} }} = \frac{{f(x)}}{{\sqrt {{x^2} + 1} }}\)
Mặt khác: \(f(x) > x + \sqrt {{x^2}} = x + \left| x \right| \ge 0,{\rm{ }}\forall x \in \mathbb{R}\)
Nên \(2xf'(x) - f(x) \ge 0 \Leftrightarrow \frac{{2xf(x)}}{{\sqrt {{x^2} + 1} }} - f(x) \ge 0\)
\( \Leftrightarrow 2x \ge \sqrt {{x^2} + 1} \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\3{x^2} \ge 1\end{array} \right. \Leftrightarrow x \ge \frac{1}{{\sqrt 3 }}\).
A. \( - 2 \le x \le \sqrt 2 \)
B. \(x \le \sqrt 2 \)
C. \( - 2 \le x\)
D. x < 0
TXĐ: \(D = \left[ { - 2;2} \right]\)
Ta có: \(f'(x) = 1 - \frac{x}{{\sqrt {4 - {x^2}} }} \Rightarrow f'(x) > 0 \Leftrightarrow \sqrt {4 - {x^2}} > x\)
\( \Leftrightarrow \left[ \begin{array}{l} - 2 \le x < 0\\\left\{ \begin{array}{l}x \ge 0\\4 - {x^2} > {x^2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 \le x < 0\\0 \le x < \sqrt 2 \end{array} \right. \Leftrightarrow - 2 \le x < \sqrt 2 \).
Ta có: \(f'(x) = 1 - \frac{x}{{\sqrt {4 - {x^2}} }} \Rightarrow f'(x) > 0 \Leftrightarrow \sqrt {4 - {x^2}} > x\)
\( \Leftrightarrow \left[ \begin{array}{l} - 2 \le x < 0\\\left\{ \begin{array}{l}x \ge 0\\4 - {x^2} > {x^2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 \le x < 0\\0 \le x < \sqrt 2 \end{array} \right. \Leftrightarrow - 2 \le x < \sqrt 2 \).
A. (- ∞; 1)\{ - 1; 0}.
B. ( - 1; + ∞)
C. (- ∞; 1)
D. ( - 1; +∞)
Chọn A
$f'\left( x \right) > 0 \Leftrightarrow \frac{{ - x + 1}}{{2\sqrt x .{{\left( {x + 1} \right)}^2}}} > 0 \Leftrightarrow \left\{ \begin{array}{l} - x + 1 > 0\\ x \ne 0\\ x \ne - 1 \end{array} \right.$
$f'\left( x \right) > 0 \Leftrightarrow \frac{{ - x + 1}}{{2\sqrt x .{{\left( {x + 1} \right)}^2}}} > 0 \Leftrightarrow \left\{ \begin{array}{l} - x + 1 > 0\\ x \ne 0\\ x \ne - 1 \end{array} \right.$
A. $( - \infty ;\sqrt {\frac{1}{2}} {\rm{]}}$
B. ${\rm{[}}\sqrt {\frac{1}{2}} ;\, + \infty ).$
C. $( + \infty ;\,\sqrt[3]{{\frac{1}{2}}}\,{\rm{]}}$
D. ${\rm{[}}{\mkern 1mu} {\kern 1pt} \sqrt[3]{{\frac{1}{2}}}{\mkern 1mu} {\kern 1pt} ; + \infty )$
Chọn D
$f'\left( x \right) \le 0 \Leftrightarrow \frac{{ - 2{x^3} + 1}}{{{{\left( {{x^3} + 1} \right)}^2}}} \le 0 \Leftrightarrow \left\{ \begin{array}{l} - 2{x^3} + 1 \le 0\\ x \ne 0 \end{array} \right. \Leftrightarrow x \ge \sqrt[3]{{\frac{1}{2}}}$
$f'\left( x \right) \le 0 \Leftrightarrow \frac{{ - 2{x^3} + 1}}{{{{\left( {{x^3} + 1} \right)}^2}}} \le 0 \Leftrightarrow \left\{ \begin{array}{l} - 2{x^3} + 1 \le 0\\ x \ne 0 \end{array} \right. \Leftrightarrow x \ge \sqrt[3]{{\frac{1}{2}}}$
Sửa lần cuối: