Question

Найти производную:
$1)y = \frac{ \sin \alpha - \cos \alpha }{ \sin \alpha + \cos \alpha } \\ 2)y = xarcos \sqrt{x}$
$3)y = 2 ln\cos(x)\times (1 - {x}^{2} )$
$4)y = {sin}^{2} (xcos {x}^{2} )$

Answer 1

$1)\; \; y=\frac{sina-cosa}{sina+cosa}\\\\y'=\frac{(cosa+sina)^2-(sina-cosa)\cdot (cosa-sina)}{(sina+cosa)^2}=\frac{(cosa+sina)^2+(sina-cosa)^2}{(sina+cosa)^2}=\\\\=\frac{cos^2a+sin^2a+2cosa\cdot sina+sin^2a+cos^2a-2cosa\cdot sina}{(sina+cosa)^2}=\frac{2}{(sina+cosa)^2}=\frac{2}{1+2sin2a}\; ;\\\\2)\; \; y=x\cdot arccos\sqrt{x}\\\\y'=1\cdot arccos\sqrt{x}+x\cdot \frac{-1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}=arccos\sqrt{x}-\frac{x}{2\sqrt{x(1-x)}}\; ;\\\\3)\; \; y=2\cdot ln(cosx)\cdot (1-x^2)$

$y'=2\cdot \frac{-sinx}{cosx}\cdot (1-x^2)+2\, ln(cosx)\cdot (-2x)=-2\, tgx\cdot (1-x^2)-4x\cdot ln(cosx)\\\\4)\; \; y=sin^2(x\, cosx^2)\\\\y'=2sin(x\, cosx^2)\cdot cos(x\, cosx^2)\cdot (1\cdot cosx^2-x\cdot sinx^2\cdot 2x)=\\\\=sin(2x\, cosx^2)\cdot (cosx^2-2x^2\cdot sinx^2)$