Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y = \frac{1}{{{{\cos }^2}3x}} = {\cos ^{ - 2}}3x\\
\Rightarrow y' = \left( { - 2} \right).\left( {\cos 3x} \right)'.co{s^{ - 3}}3x = - 2.\left( {3x} \right)'.\left( { - \sin 3x} \right).{\cos ^{ - 3}}3x = \frac{{ - 6\sin 3x}}{{{{\cos }^3}3x}}\\
b,\\
y = \frac{{\sin x - \cos x}}{{\sin x + \cos x}}\\
\Rightarrow y' = \frac{{\left( {\sin x - \cos x} \right)'.\left( {\sin x + \cos x} \right) - \left( {\sin x + \cos x} \right)'.\left( {\sin x - \cos x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}}\\
= \frac{{\left( {\cos x + \sin x} \right)\left( {\sin x + \cos x} \right) - \left( {\cos x - \sin x} \right)\left( {\sin x - \cos x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}}\\
= \frac{{\left( {{{\sin }^2}x + 2\sin x.\cos x + {{\cos }^2}x} \right) + \left( {{{\sin }^2}x - 2\sin x.\cos x + {{\cos }^2}x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}}\\
= \frac{{2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}} = \frac{2}{{{{\left( {\sin x + \cos x} \right)}^2}}}\\
c,\\
y = {\tan ^3}\left( {x + \pi } \right) - 4{x^2} + \sqrt x = {\tan ^3}x - 4{x^2} + \sqrt x \\
\Rightarrow y' = 3.\left( {\tan x} \right)'.{\tan ^2}x - 4.2x + \frac{1}{{2\sqrt x }}\\
= 3.\frac{1}{{{{\cos }^2}x}}.{\tan ^2}x - 8x + \frac{1}{{2\sqrt x }}\\
= \frac{{3{{\tan }^2}x}}{{{{\cos }^2}x}} - 8x + \frac{1}{{2\sqrt x }}\\
d,\\
\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\cos \left( {\frac{\pi }{2} - x} \right)}}{{\sin \left( {\frac{\pi }{2} - x} \right)}} = \cot \left( {\frac{\pi }{2} - x} \right)\\
y = 7{\cot ^4}\left( {3x - \frac{\pi }{2}} \right) + 3\sin x = 7{\tan ^4}x + 3\sin x\\
\Rightarrow y' = 7.4.\left( {\tan x} \right)'.{\tan ^3}x + 3\cos x\\
= 28.\frac{1}{{{{\cos }^2}x}}.{\tan ^3}x + 3\cos x\\
= \frac{{28{{\tan }^3}x}}{{{{\cos }^2}x}} + 3\cos x
\end{array}\)
