Question

$f(x)=cos^{2}(x)*tg^{2}(x)$
$f'(\frac{\pi}{4})=?$

Answer 1

Ответ:

$f'\Big(\frac{\pi}{4}\Big)=1$

Пошаговое объяснение:

$f(x)=cos^2(x) \cdot tg^2(x)=cos^2(x) \cdot \frac{sin^2(x)}{cos^2(x)}=sin^2(x)\\ \\\\f'(x)=(sin^2(x))'=2sin(x)(sin(x))'=2sin(x)cos(x)=sin(2x)\\ \\\\f'\Big(\frac{\pi}{4}\Big)=sin\big(\frac{\pi}{4} \cdot2\big)=sin\big(\frac{\pi}{2}\big)=1$