Các quy tắc tính đạo hàm

\({(2\cos x)^\prime } = 2.\left( {\cos x} \right)' \)\(=2.(-\sin x)=  - 2\sin x\)

\(y' = \dfrac{1}{{{{\cos }^2}x}}\)

B sai vì \(\left( {\sqrt x } \right)' = \dfrac{1}{{2\sqrt x }}\).

\(y = {x^4} - 3{x^2} + 2x - 1 \Rightarrow y' = 4{x^3} - 3.2x + 2 = 4{x^3} - 6x + 2\)

\(y' = \dfrac{{\left( {2x + 1} \right)'.\left( {x + 2} \right) - \left( {2x + 1} \right)\left( {x + 2} \right)'}}{{{{\left( {x + 2} \right)}^2}}} = \dfrac{{2\left( {x + 2} \right) - 2x - 1}}{{{{\left( {x + 2} \right)}^2}}} = \dfrac{3}{{{{\left( {x + 2} \right)}^2}}}\)

\(f'\left( x \right) = 9{x^2} + 2x;y' \le 0\) \( \Leftrightarrow 9{x^2} + 2x \le 0 \Leftrightarrow  - \dfrac{2}{9} \le x \le 0\)

Vậy \(S = \left[ { - \dfrac{2}{9};0} \right]\)

\(f\left( x \right) = \sqrt[3]{x} = {x^{\frac{1}{3}}} \Rightarrow f'\left( x \right) = \dfrac{1}{3}.{x^{\frac{1}{3} - 1}} = \dfrac{1}{3}{x^{ - \frac{2}{3}}} = \dfrac{1}{3}\dfrac{1}{{{x^{\frac{2}{3}}}}} = \dfrac{1}{3}\dfrac{1}{{\sqrt[3]{{{x^2}}}}}\) \(\Rightarrow f'\left( 8 \right) = \dfrac{1}{3}.\dfrac{1}{{\sqrt[3]{{{8^2}}}}} = \dfrac{1}{{12}}\)

Bước 1:

\(y' = \dfrac{{3'\left( {1 - x} \right) - 3\left( {1 - x} \right)'}}{{{{\left( {1 - x} \right)}^2}}} = \dfrac{{ - 3.\left( { - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}} \)\(= \dfrac{3}{{{{\left( {1 - x} \right)}^2}}}  \)

Bước 2:

Ta có \(y'=\dfrac{3}{{{{\left( {1 - x} \right)}^2}}} > 0\,\,\forall x \ne 1 \)

\(\Rightarrow \)Tập nghiệm của bất phương trình \(y' < 0\) là \(\emptyset \).

Đáp án A: \(y' = \dfrac{{\left( {{x^3} + 1} \right)'.x - \left( {{x^3} + 1} \right)x'}}{{{x^2}}} \) \(= \dfrac{{3{x^2}.x - {x^3} - 1}}{{{x^2}}} = \dfrac{{2{x^3} - 1}}{{{x^2}}}\)

Đáp án B:

\(\begin{array}{l}y = \dfrac{{3\left( {x + 1} \right)}}{{{x^2}}}\\ \Rightarrow y' = 3.\dfrac{{\left( {x + 1} \right)'.{x^2} - \left( {x + 1} \right)\left( {{x^2}} \right)'}}{{{x^4}}}\\ = 3\dfrac{{{x^2} - 2x\left( {x + 1} \right)}}{{{x^4}}}\\ = 3\dfrac{{ - {x^2} - 2x}}{{{x^4}}} =  - 3\dfrac{{x + 2}}{{{x^3}}}\end{array}\)

Đáp án C: \(y' = \dfrac{{\left( {{x^3} + 5x - 1} \right)'.x - \left( {{x^3} + 5x - 1} \right).x'}}{{{x^2}}} \) \(= \dfrac{{\left( {3{x^2} + 5} \right).x - {x^3} - 5x + 1}}{{{x^2}}} \) \(= \dfrac{{2{x^3} + 1}}{{{x^2}}} = 2x + \dfrac{1}{{{x^2}}}\)

\(\begin{array}{l}y = \dfrac{1}{{{x^3}}} - \dfrac{1}{{{x^2}}} = {x^{ - 3}} - {x^{ - 2}}\\ \Rightarrow y' =  - 3{x^{ - 4}} - \left( { - 2} \right){x^{ - 3}} = \dfrac{{ - 3}}{{{x^4}}} + \dfrac{2}{{{x^3}}}\end{array}\)

\(\begin{array}{l}y' = \dfrac{{\left( {ax + b} \right)'\left( {cx + d} \right) - \left( {ax + b} \right)\left( {cx + d} \right)'}}{{{{\left( {cx + d} \right)}^2}}} = \dfrac{{a\left( {cx + d} \right) - c\left( {ax + b} \right)}}{{{{\left( {cx + d} \right)}^2}}}\\ = \dfrac{{acx + ad - acx - bc}}{{{{\left( {cx + d} \right)}^2}}} = \dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\end{array}\)

\(\begin{array}{l}y' = \dfrac{{\left( {{x^2} - x + 1} \right)'\left( {x - 1} \right) - \left( {{x^2} - x + 1} \right)\left( {x - 1} \right)'}}{{{{\left( {x - 1} \right)}^2}}} = \dfrac{{\left( {2x - 1} \right)\left( {x - 1} \right) - \left( {{x^2} - x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\ = \dfrac{{2{x^2} - 2x - x + 1 - {x^2} + x - 1}}{{{{\left( {x - 1} \right)}^2}}} = \dfrac{{{x^2} - 2x}}{{{{\left( {x - 1} \right)}^2}}}\end{array}\)

\(\begin{array}{l}y = {\left( {{x^7} + x} \right)^2} = {x^{14}} + 2{x^8} + {x^2}\\ \Rightarrow y' = 14{x^{13}} + 2.8{x^7} + 2x\\ = 14{x^{13}} + 16{x^7} + 2x\\ = 2\left( {7{x^{13}} + 8{x^7} + x} \right)\\ = 2\left( {7{x^{13}} + 7{x^7} + {x^7} + x} \right) \\= 2\left[ {7{x^6}\left( {{x^7} + x} \right) + {x^7} + x} \right]\\= 2\left( {{x^7} + x} \right)\left( {7{x^6} + 1} \right)\end{array}\)

\(\begin{array}{l}y = \dfrac{1}{{x\sqrt x }} = \dfrac{1}{{x.{x^{\frac{1}{2}}}}} = \dfrac{1}{{{x^{1 + \frac{1}{2}}}}} = \dfrac{1}{{{x^{\frac{3}{2}}}}} = {x^{ - \frac{3}{2}}}\\ \Rightarrow y' =  - \dfrac{3}{2}{x^{ - \frac{3}{2} - 1}} =  - \dfrac{3}{2}{x^{ - \frac{5}{2}}} =  - \dfrac{3}{2}\dfrac{1}{{{x^{\frac{5}{2}}}}} =  - \dfrac{3}{2}\dfrac{1}{{{x^2}\sqrt x }}\end{array}\)

Bước 1:

\(\begin{array}{l}y = \sin 2x = 2\sin x\cos x\\ \Rightarrow y' = \left( {2\sin x\cos x} \right)'\\ = 2\left( {\sin x\cos x} \right)'\\ = 2\left[ {\left( {\sin x} \right)'.\cos x + \sin x.\left( {\cos x} \right)'} \right]\end{array}\)

Bước 2:

$= 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\\ = 2\cos 2x$

\(\begin{array}{l}y' = \dfrac{{\left( {2{x^2} + 3x - 1} \right)'\left( {{x^2} - 5x + 2} \right) - \left( {2{x^2} + 3x - 1} \right)\left( {{x^2} - 5x + 2} \right)'}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\\y' = \dfrac{{\left( {4x + 3} \right)\left( {{x^2} - 5x + 2} \right) - \left( {2{x^2} + 3x - 1} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\\y' = \dfrac{{4{x^3} - 20{x^2} + 8x + 3{x^2} - 15x + 6 - 4{x^3} - 6{x^2} + 2x + 10{x^2} + 15x - 5}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\\y' = \dfrac{{ - 13{x^2} + 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\end{array}\)

Có: \(f'\left( x \right) = 3{x^2} - 3.2x = 3{x^2} - 6x\)

\(f'\left( x \right) < 0 \Leftrightarrow 3{x^2} - 6x < 0 \Leftrightarrow 0 < x < 2\)

\(\begin{array}{l}f\left( x \right) = {\left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right)^3} = {\left( {\sqrt x } \right)^3} - 3{\left( {\sqrt x } \right)^2}.\dfrac{1}{{\sqrt x }} + 3\sqrt x {\left( {\dfrac{1}{{\sqrt x }}} \right)^2} - {\left( {\dfrac{1}{{\sqrt x }}} \right)^3}\\f\left( x \right) = {x^{\dfrac{3}{2}}} - 3\sqrt x  + \dfrac{3}{{\sqrt x }} - \dfrac{1}{{{x^{\dfrac{3}{2}}}}}\\f\left( x \right) = {x^{\dfrac{3}{2}}} - 3\sqrt x  + 3{x^{ - \dfrac{1}{2}}} - {x^{ - \dfrac{3}{2}}}\\f'\left( x \right) = \dfrac{3}{2}{x^{\dfrac{3}{2} - 1}} - \dfrac{3}{{2\sqrt x }} + 3.\left( { - \dfrac{1}{2}} \right){x^{ - \dfrac{1}{2} - 1}} + \dfrac{3}{2}{x^{ - \dfrac{3}{2} - 1}}\\f'\left( x \right) = \dfrac{3}{2}\sqrt x  - \dfrac{3}{{2\sqrt x }} - \dfrac{3}{2}{x^{ - \dfrac{3}{2}}} + \dfrac{3}{2}{x^{ - \dfrac{5}{2}}}\\f'\left( x \right) = \dfrac{3}{2}\left( {\sqrt x  - \dfrac{1}{{\sqrt x }} - \dfrac{1}{{x\sqrt x }} + \dfrac{1}{{{x^2}\sqrt x }}} \right)\end{array}\)

$\begin{array}{l}y = {\tan ^2}x - co{t^2}x = \left( {\tan x - \cot x} \right)\left( {\tan x + \cot x} \right)\\y' = \left( {\tan x - \cot x} \right)'\left( {\tan x + \cot x} \right) + \left( {\tan x - \cot x} \right)\left( {\tan x + \cot x} \right)'\\y' = \left( {\dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}} \right)\left( {\tan x + \cot x} \right) + \left( {\tan x - \cot x} \right)\left( {\dfrac{1}{{{{\cos }^2}x}} - \dfrac{1}{{{{\sin }^2}x}}} \right)\\y' = \dfrac{{\tan x}}{{{{\cos }^2}x}} + \dfrac{{\cot x}}{{{{\cos }^2}x}} + \dfrac{{\tan x}}{{{{\sin }^2}x}} + \dfrac{{\cot x}}{{{{\sin }^2}x}} + \dfrac{{\tan x}}{{{{\cos }^2}x}} - \dfrac{{\tan x}}{{{{\sin }^2}x}} - \dfrac{{\cot x}}{{{{\cos }^2}x}} + \dfrac{{\cot x}}{{{{\sin }^2}x}}\\y' = 2\dfrac{{\tan x}}{{{{\cos }^2}x}} + 2\dfrac{{\cot x}}{{{{\sin }^2}x}}\end{array}$

\(\begin{array}{l}f\left( x \right) = \tan \left( {x - \dfrac{{2\pi }}{3}} \right) = \dfrac{{\tan x - \tan \dfrac{{2\pi }}{3}}}{{1 + \tan x.\tan \dfrac{{2\pi }}{3}}} = \dfrac{{\tan x + \sqrt 3 }}{{1 - \sqrt 3 \tan x}}\\f'\left( x \right) = \dfrac{{\left( {\tan x + \sqrt 3 } \right)'\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan x} \right)'}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{{\dfrac{1}{{{{\cos }^2}x}}\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( { - \dfrac{{\sqrt 3 }}{{{{\cos }^2}x}}} \right)}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{{\dfrac{1}{{{{\cos }^2}x}} - \dfrac{{\sqrt 3 \tan x}}{{{{\cos }^2}x}} + \dfrac{{\sqrt 3 \tan x}}{{{{\cos }^2}x}} + \dfrac{3}{{{{\cos }^2}x}}}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{4}{{{{\cos }^2}x{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\ \Rightarrow f'\left( 0 \right) = \dfrac{4}{{1\left( {1 - \sqrt 3 .0} \right)}} = 4\end{array}\)