Question

Care ma ajuta sa calculez derivata funcției x/1+x^2

Answer 1

$y = \frac{x}{1 + x^{2}} = x \cdot (1 + x^2)^{-1}\\y^{'} = (\frac{d}{dx}[x]\cdot (1+x^{2})^{-1}) + x \cdot \frac{d}{dx} [(1+x^{2})^{-1}]\\= \frac{1}{1+x^{2}} + x \cdot (-1)(1+x^2)^{-2}(2x) = \boxed{\frac{1}{1 + x^2} - \frac{2x^2}{(1+x^2)^2}}$