Question

Знатоки, хелп!! Нужна помощь, срочно :((​

Answer 1

Ответ:

а) f'(x)=cos x+2sin x

b) $f'(\frac{\pi}{6})=2\frac{5}{6}$

c) $f'(x)=\frac{\sqrt{2}}{2}+4$

d) $f'(\frac{\pi}{3})=-\frac{5}{6}$

Объяснение:

а) f'(x)=cos x+2sin x

b) $f'(x)=\frac{2}{\cos^2x}+\frac{1}{3}\sin x$

$f'(\frac{\pi}{6})=\frac{2}{\cos^2\frac{\pi}{6}}+\frac{\sin\frac{\pi}{6}}{3}$

$f'(\frac{\pi}{6})=\frac{2}{(\frac{\sqrt{3}}{2})^2}+\frac{\frac{1}{2}}{3}$

$f'(\frac{\pi}{6})=\frac{2}{\frac{3}{4}} +\frac{1}{6}$

$f'(\frac{\pi}{6})=\frac{8}{3}+\frac{1}{6}$

$f'(\frac{\pi}{6})=2\frac{2}{3}+\frac{1}{6}$

$f'(\frac{\pi}{6})=2\frac{5}{6}$

c) $f'(x)=\cos x+\frac{1}{\cos^2x}+\frac{8x}{\pi}$

$f'(x)=\cos\frac{\pi}{4}+\frac{1}{\cos^2\frac{\pi}{4}}+\frac{8\frac{\pi}{4}}{\pi}$

$f'(x)=\frac{\sqrt{2}}{2}+\frac{1}{(\frac{\sqrt{2}}{2})^2}+\frac{8}{4}$

$f'(x)=\frac{\sqrt{2}}{2}+\frac{1}{\frac{1}{2}}+2$

$f'(x)=\frac{\sqrt{2}}{2}+2+2$

$f'(x)=\frac{\sqrt{2}}{2}+4$

d) $f(x)=\sqrt{3}\cos x+\sin\frac{\pi}{6}+\frac{x^2}{\pi}$

$f'(x)=-\sqrt{3}\sin x+\frac{2x}{\pi}$

$f'(\frac{\pi}{3})=-\sqrt{3}\sin(\frac{\pi}{3})+\frac{2(\frac{\pi}{3})}{\pi}$

$f'(\frac{\pi}{3})=-\sqrt{3}\sin(\frac{\pi}{3})+\frac{2}{3}$

$f'(\frac{\pi}{3})=-\sqrt{3}\frac{\sqrt{3}}{2}+\frac{2}{3}$

$f'(\frac{\pi}{3})=-\frac{3}{2}+\frac{2}{3}$

$f'(\frac{\pi}{3})=-\frac{5}{6}$