Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = {\sin ^2}\left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
\Rightarrow y' = 2.\left[ {\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)} \right]'.\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
= 2.\left( {\cos \left( {{{\tan }^4}3x} \right)} \right)'.\cos \left( {\cos \left( {{{\tan }^4}3x} \right)} \right).\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
= 2.\left( {{{\tan }^4}3x} \right)'.\left( { - \sin \left( {{{\tan }^4}3x} \right)} \right).\cos \left( {\cos \left( {{{\tan }^4}3x} \right)} \right).\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
= - 2.4.\left( {\tan 3x} \right)'.{\tan ^3}3x.\sin \left( {{{\tan }^4}3x} \right).\cos \left( {\cos \left( {{{\tan }^4}3x} \right)} \right).\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
= - 8.\frac{{\left( {3x} \right)'}}{{{{\cos }^2}3x}}.{\tan ^3}3x.\sin \left( {{{\tan }^4}3x} \right).\cos \left( {\cos \left( {{{\tan }^4}3x} \right)} \right).\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)\\
= \frac{{ - 24}}{{{{\cos }^3}3x}}.{\tan ^3}3x.\sin \left( {{{\tan }^4}3x} \right).\cos \left( {\cos \left( {{{\tan }^4}3x} \right)} \right).\sin \left( {\cos \left( {{{\tan }^4}3x} \right)} \right)
\end{array}\)
