Question

Найти производные функций

Answer 1

$1)\; \; y=arccos\dfrac{cosx}{\sqrt{1+cos^2x}}\\\\\\y'=-\dfrac{1}{\sqrt{1-\frac{cos^2x}{1+cos^2x}}}\cdot \dfrac{-sinx\cdot \sqrt{1+cos^2x}-cosx\cdot \frac{1}{2\sqrt{1+cos^2x}}\cdot (-2cosx\cdot sinx)}{1+cos^2x}=\\\\\\=\sqrt{1+cos^2x}\cdot \dfrac{sinx\cdot (1+cos^2x)-cos^2x\cdot sinx}{\sqrt{1+cos^2x}\cdot (1+cos^2x)}=\dfrac{sinx}{1+cos^2x}$

$2)\; \; y=(arcctgx)^{x}\; \; \; \Rightarrow \quad lny=x\cdot ln(arcctgx)\\\\\\\dfrac{y'}{y}=ln(atcctgx)+x\cdot \dfrac{1}{arcctgx}\cdot \dfrac{-1}{1+x^2}=ln(arcctgx)-\dfrac{x}{(1+x^2)\, arcctgx}\\\\\\y'=(arcctgx)^{x}\cdot \left (ln(arcctgx)-\dfrac{x}{(1+x^2)\, arcctgx}\right)$

$3)\; \; x^2+y^3-3xy=0\; \; ,\; \; y=y(x)\\\\2x+3y^2\cdot y'-(3y+3x\, y')=0\\\\2x+3y^2\cdot y'-3y-3x\, y'=0\\\\y'(3y^2-3x)=3y-2x\\\\y'=\dfrac{3y-2x}{3(y^2-x)}$