Question

ПРОИЗВОДНАЯ
11 КЛАСС

Answer 1

$f'(x)=(frac{cos4x^2}{sqrtpi})'=frac{-sin4x^2cdot(4x^2)'}{sqrtpi}=frac{-8xsin4x^2}{sqrtpi}\f'(x_0)=frac{-8x_0sin4x_0^2}{sqrtpi}=-frac{8sqrtpisinpi}{2sqrtpi}=0$