Question

Найти производную. пожжалуйста помогите y=sqrt(2cos(x))+sin*(p/4)+(2/p)*x^2

Answer 1

$y=\sqrt{2cosx}+\underbrace {sin\frac{\pi}{4}}_{const}+\frac{2}{\pi}{x^2\; ,\; \; x_0=\frac{\pi}{4}$

$y'= \frac{1}{2\sqrt{2cosx}}\cdot (2cosx)'+\frac{2}{\pi }\cdot 2x=\frac{1}{2\sqrt{2cosx}}\cdot (-2sinx)+ \frac{4x}{\pi }=\\\\=-\frac{sinx}{\sqrt{2cosx}} +\frac{4x}{\pi }\\\\y'(\frac{\pi}{4})= -\frac{sin\frac{\pi}{4}}{\sqrt{2cos\frac{\pi}{4}}}+\frac{4\cdot \frac{\pi}{4}}{\pi}=- \frac{\frac{\sqrt2}{2}}{\sqrt{2\cdot \frac{\sqrt2}{2}}}+1=- \frac{\sqrt2}{2\sqrt[4]2}+1=1- \frac{\sqrt[4]2}{2}$