Đáp án:

Giải thích các bước giải:

25b

26

27b

28 đáp án là 4/ (x-2)^3

29 đáp án là (2 sin x)/(cosx)^3

30c

31d

32d

33a

34a

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
25,\\
y = \cot x \Rightarrow y' = \dfrac{{ - 1}}{{{{\sin }^2}x}}\\
26,\\
y = \sqrt {\cot 2x}  = \dfrac{{\left( {\cot 2x} \right)'}}{{2\sqrt {\cot 2x} }} = \dfrac{{\dfrac{{ - \left( {2x} \right)'}}{{{{\sin }^2}2x}}}}{{2\sqrt {\cot 2x} }}\\
 = \dfrac{{ - \dfrac{2}{{{{\sin }^2}2x}}}}{{2\sqrt {\cot 2x} }} = \dfrac{{\dfrac{{ - 1}}{{{{\sin }^2}2x}}}}{{\sqrt {\cot 2x} }} = \dfrac{{ - \left( {1 + {{\cot }^2}2x} \right)}}{{\sqrt {\cot 2x} }}\\
27,\\
y = \cos 3x.sin2x\\
 \Rightarrow y' = \left( {\cos 3x} \right)'.sin2x + \cos 3x.\left( {\sin 2x} \right)'\\
 = \left( {3x} \right)'.\left( { - \sin 3x} \right).sin2x + \cos 3x.\left( {2x} \right)'.cos2x\\
 =  - 3\sin 3x.sin2x + 2\cos 3x.cos2x\\
 \Rightarrow y'\left( {\dfrac{\pi }{3}} \right) = 1\\
28,\\
y = \dfrac{x}{{x - 2}}\\
 \Rightarrow y' = \dfrac{{x'.\left( {x - 2} \right) - \left( {x - 2} \right)'.x}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{1.\left( {x - 2} \right) - 1.x}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{ - 2}}{{{{\left( {x - 2} \right)}^2}}}\\
 \Rightarrow y'' = \dfrac{{ - 2. - \left[ {{{\left( {x - 2} \right)}^2}} \right]'}}{{{{\left( {x - 2} \right)}^4}}} = \dfrac{{2.2.\left( {x - 2} \right)'.\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^4}}} = \dfrac{4}{{{{\left( {x - 2} \right)}^3}}}\\
29,\\
y = \tan x\\
 \Rightarrow y' = \dfrac{1}{{{{\cos }^2}x}}\\
 \Rightarrow y'' = \dfrac{{ - \left( {{{\cos }^2}x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{ - 2.\left( {\cos x} \right)'.\cos x}}{{{{\cos }^4}x}} = \dfrac{{ - 2\left( { - \sin x} \right).\cos x}}{{{{\cos }^4}x}} = \dfrac{{2\sin x}}{{{{\cos }^3}x}}\\
30,\\
y = {\left( {{x^4} - 1} \right)^3}\\
 \Rightarrow y' = 3.\left( {{x^4} - 1} \right).{\left( {{x^4} - 1} \right)^2} = 3.4{x^3}.{\left( {{x^4} - 1} \right)^2} = 12{x^3}.{\left( {{x^4} - 1} \right)^2}\\
31,\\
f\left( x \right) = \sqrt {{x^2} - 2x + 4} \\
 \Rightarrow f'\left( x \right) = \dfrac{{\left( {{x^2} - 2x + 4} \right)'}}{{2\sqrt {{x^2} - 2x + 4} }} = \dfrac{{2x - 2}}{{2\sqrt {{x^2} - 2x + 4} }} = \dfrac{{x - 1}}{{\sqrt {{x^2} - 2x + 4} }}\\
32,\\
y = \dfrac{{2x - 3}}{{x + 4}}\\
 \Rightarrow y' = \dfrac{{\left( {2x - 3} \right)'.\left( {x + 4} \right) - \left( {x + 4} \right)'.\left( {2x - 3} \right)}}{{{{\left( {x + 4} \right)}^2}}}\\
 = \dfrac{{2.\left( {x + 4} \right) - 1.\left( {2x - 3} \right)}}{{{{\left( {x + 4} \right)}^2}}} = \dfrac{7}{{{{\left( {x + 4} \right)}^2}}}\\
33,\\
y = \left( {{x^3} - 2x} \right)\left( {x + 3} \right) = {x^4} + 3{x^3} - 2{x^2} - 6x\\
 \Rightarrow y' = 4{x^3} + 9{x^2} - 4x - 6\\
34,\\
y = {x^3} + 2{x^2} - 4x + 5\\
 \Rightarrow y' = 3{x^2} + 4x - 4
\end{array}\)