Question

530.Посчитайте $f'(a)$
c)$f(x)=sin(2x),a=\frac{\pi }{2}$
d)$f(x)=2tg(x)-sin(5),a=\frac{\pi }{4}$

Answer 1

Ответ:

$1)\ \ f(x)=sin(2x)\\\\\star \ \ (sinu)'=cosu\cdot u'\ \ ,\ \ u=2x\ \ \star \\\\f'(x)=cos2x\cdot (2x)'=cos2x\cdot 2=2\, cos2x\\\\f'(\dfrac{\pi }{2})=2\cdot cos\pi =2\cdot (-1)=-2\\\\\\2)\ \ f(x)=2\, tgx-\underbrace {sin5}_{const}\\\\f'(x)=2\cdot (tgx)'-\underbrace{(sin5)'}_{C'=0}=2\cdot \dfrac{1}{cos^2x}\\\\f'(\dfrac{\pi}{4})=2\cdot \dfrac{1}{cos^2\frac{\pi}{4}}=2\cdot \dfrac{1}{(\frac{\sqrt2}{2})^2}=\dfrac{2}{\frac{2}{4}}=4$