Question

Вычислите значение производной функции у = 1\2tg(4x-p)+p\4 x0=p\4

Answer 1

$y=\frac{1}{tg(4x-\pi )+\frac{\pi}{4}}\quad \quad (\frac{1}{u})'=-\frac{1}{u^2}\cdot u'$

$y'=-\frac{1}{(tg(4x-\pi )+\frac{\pi}{4})^2}\cdot \frac{4}{cos^2(4x-\pi )}\\\\y'(\frac{\pi}{4})=-\frac{1}{(tg(4\cdot \frac{\pi}{4}-\pi )+\frac{\pi}{4})^2}\cdot \frac{4}{cos^20}=-\frac{1}{(0+\frac{\pi}{4})^2}\cdot \frac{4}{1}=-\frac{64}{\pi ^2}$

$2)y=\frac{1}{2}\cdot tg(4x-\pi)+\frac{\pi}{4}\\\\y'=\frac{1}{2}\cdot\frac{4}{cos^2(4x-\pi)}\\\\y'(\frac{\pi}{4})=\frac{2}{cos^20}=\frac{2}{1}=2$