Đáp án:
\[y' = - 20.\sin 4x.{\cos ^{18}}2x\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1\\
{\cos ^2}x - {\sin ^2}x = {\cos ^2}x - \left( {1 - {{\cos }^2}x} \right) = 2{\cos ^2}x - 1 = \cos 2x\\
y = {\left( {\frac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} \right)^{20}} = {\left( {\frac{{1 - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}{{1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}} \right)^{20}} = {\left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x + {{\sin }^2}x}}} \right)^{20}} = {\left( {{{\cos }^2}x - {{\sin }^2}x} \right)^{20}} = {\left( {\cos 2x} \right)^{20}}\\
\Rightarrow y' = 20.\left( {\cos 2x} \right)'.{\left( {\cos 2x} \right)^{19}} = 20.2.\left( { - \sin 2x} \right).{\cos ^{19}}2x = - 20.\left( {2\sin 2x.\cos 2x} \right).{\cos ^{18}}2x = - 20.\sin 4x.{\cos ^{18}}2x
\end{array}\)
