Помогите найти производные)19задание, заранее спасибо)

Решите задачу:

$1)\quad y=7\sqrt[5]{x^3-2x+\frac{2}{x}}\\\\y'=7\cdot \frac{1}{2\sqrt{x^3-2x+\frac{2}{x}}}\cdot (3x^2-2-\frac{2}{x^2})\\\\2)\quad y=3^{x}\cdot e^{-x^2}\\\\y'=3^{x}\cdot ln3\cdot e^{-x^2}+3^{x}\cdot e^{-x^2}\cdot (-2x)\\\\3)\quad y= \frac{arccosx}{\sqrt{1-x^2}} \\\\y'=\frac{-\frac{1}{\sqrt{1-x^2}}\cdot \sqrt{1-x^2}-arccosx\cdot \frac{-2x}{2\sqrt{1-x^2}}}{1-x^2}= \frac{-\sqrt{1-x^2}+x\cdot arccosx}{\sqrt{(1-x^2)^3}}$

$4)\quad ln( y+1)=arctg \frac{x}{y} \\\\ \frac{y'}{y+1} =\frac{1}{1+\frac{x^2}{y^2}}\cdot \frac{y-xy'}{y^2}\\\\\frac{y'}{y+1}= \frac{y^2}{x^2+y^2} \cdot \frac{y-xy'}{y^2} \\\\\frac{y'}{y+1}= \frac{y-xy'}{x^2+y^2} \\\\y'(x^2+y^2)=(y+1)(y-xy')\\\\y'(x^2+y^2)=y(y+1)-xy'(y+1)\\\\y'(x^2+y^2)+xy'(y+x)=y(y+1)\\\\y'(x^2+y^2+xy+x^2)=y(y+1)\\\\y'= \frac{y(y+1)}{2x^2+y^2+xy}$