Question

Найдите:
$y=arcctg \frac{1}{1+ x^{2} } ; y'=?$

Answer 1

$y=arctg\frac{1}{1+x^2}\\\\y'=\frac{1}{1+(\frac{1}{1+x^2})^2}*(\frac{1}{1+x^2})'=\frac{1}{1+\frac{1}{(1+x^2)^2}}*(\frac{-2x}{(1+x^2)^2})=\frac{(1+x^2)^2}{1+(1+x^2)^2}*(\frac{-2x}{(1+x^2)^2})\\\\=-\frac{2x}{1+(1+x^2)^2}$