Question

Найти производные y'(x) заданных функций:
$y=x*\sqrt\frac{1+x^{2} }{1-x^{2} }$

$\frac{1}{tg^{2}*2x }$

Answer 1

$1)\; \; y=x\cdot \sqrt{\frac{1+x^2}{1-x^2}}\\\\y'=\sqrt{\frac{1+x^2}{1-x^2}}+x\cdot \frac{1}{2\sqrt{\frac{1+x^2}{1-x^2}}}\cdot \frac{2x(1-x^2)-(1+x^2)\cdot (-2x)}{(1-x^2)^2}=\\\\=\sqrt{\frac{1+x^2}{1-x^2}}+\frac{x}{2}\cdot \sqrt{\frac{1-x^2}{1+x^2}}\cdot \frac{4x}{(1-x^2)^2}\\\\2)\; \; y=\frac{1}{tg^22x}\\\\y'=\frac{-2\, tg2x\cdot \frac{1}{cos^22x}\cdot 2}{tg^42x}=-\frac{4\, tg2x}{cos^22x\cdot tg^42x}=-\frac{4cos2x}{sin^32x}$