Giải thích các bước giải:
\(\begin{array}{l}
25,\\
f\left( x \right) = 2\sin 3x.\cos 5x\\
\Rightarrow f'\left( x \right) = 2.\left[ {\left( {\sin 3x} \right)'.\cos 5x + \sin 3x.\left( {\cos 5x} \right)'} \right]\\
= 2.\left[ {3\cos 3x.\cos 5x + \sin 3x.5.\left( { - \sin 5x} \right)} \right]\\
= 6\cos 3x.\cos 5x - 10.\sin 3x.\sin 5x\\
\Rightarrow f'\left( {\frac{\pi }{8}} \right) = - 8 - \sqrt 2 \\
26,\\
f\left( x \right) = {\sin ^4}x + {\cos ^4}x\\
\Rightarrow f'\left( x \right) = 4.\left( {\sin x} \right)'.{\sin ^3}x + 4.\left( {\cos x} \right)'.{\cos ^3}x\\
= 4.\cos x.{\sin ^3}x + 4.\left( { - \sin x} \right).{\cos ^3}x\\
= 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\\
\Rightarrow f'\left( {\frac{\pi }{8}} \right) = - 1\\
27,\\
f\left( x \right) = {\cos ^2}x - {\sin ^2}x\\
f'\left( x \right) = 2.\left( {\cos x} \right)'.\cos x - 2.\left( {\sin x} \right)'.sinx\\
= 2.\left( { - \sin x} \right).\cos x - 2.\cos x.\sin x\\
= - 4\sin x\cos x\\
\Rightarrow f'\left( {\frac{\pi }{4}} \right) = - 2
\end{array}\)
