Question

Помогите 45баллов. найти производные: y=7* 5√ (x^3) - 2x + 1/x y=(3^x) + e^(-x^2) y=arccosX/√1 - x^2 in(y+1)=arctg x/y

Answer 1

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

$y'= \frac{-\frac{1}{\sqrt{1-x^2}}\cdot (-2x)\cdot \sqrt{1-x^2}-arccosx\cdot \frac{-2x}{2\sqrt{1-x^2}}}{1-x^2}=\frac{2x\cdot \sqrt{1-x^2}+x\cdot arccosx}{\sqrt{(1-x^2)^3}}$

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