Question

Задание: продифферинцировать функции

Answer 1

$1)\ \ y=sin^2x\cdot sinx^2\\\\y'=(sin^2x)'\cdot sinx^2+sin^2x\cdot (sinx^2)'=\\\\=2\, sinx\cdot cosx\cdot sinx^2+sin^2x\cdot cosx^2\cdot 2x=\\\\=sin2x\cdot sinx^2+2x\cdot sin^2x\cdot cosx^2$

$2)\ \ y=\dfrac{2cosx}{\sqrt{cos2x}}\\\\\\y'=\dfrac{(2cosx)'\cdot \sqrt{cos2x}-2cosx\cdot (\sqrt{cos2x})'}{(\sqrt{cos2x})^2}=\\\\\\=\dfrac{-2sinx\cdot \sqrt{cos2x}-2cosx\cdot \frac{-sin2x\cdot 2}{2\sqrt{cos2x}}}{cos2x}=\dfrac{-2sinx\cdot cos2x+2cosx\cdot sin2x}{\sqrt{cos2x}\cdot cos2x}=\\\\\\=\dfrac{2\cdot (sin2x\cdot cosx-cos2x\cdot sinx)}{\sqrt{cos^32x}} =\dfrac{2\cdot sin(2x-x)}{\sqrt{cos^32x}}=\dfrac{2sinx}{\sqrt{cos^32x}}$