Question

Derivada de
$y= \frac{tgx}{1+x^2}$

Answer 1

$y = \frac{tgx}{1+x^2}$

use a regra do quociente:


(F(x)'G(x) -F(x)*G(x)')/G(x)²

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$\\ y = \frac{tgx}{1+x^2} \\ \\ y' = \frac{tgx'*(1+x^2) -tgx*(1+x^2)'}{(1+x^2)^2} \\ \\ y' = \frac{sec^2x*(1+x^2)-tgx(0+2x)}{(1+x^2)^2} \\ \\ y' = \frac{sec^2x(1+x^2)-2xtgx}{(1+x^2)^2}$

Answer 2

$\sf \displaystyle y=\frac{tgx}{1+x^2}\\\\\\\frac{d}{dx}\left(\frac{tan \left(x\right)}{1+x^2}\right)\\\\\\{Aplique\:a\:regra\:do\:quociente}\to :\quad \left(\frac{f}{g}\right)^'=\dfrac{f\:'\cdot g-g'\cdot f}{g^2}$

$\sf \displaystyle =\frac{\frac{d}{dx}\left(tan \left(x\right)\right)\left(1+x^2\right)-\frac{d}{dx}\left(1+x^2\right)tan \left(x\right)}{\left(1+x^2\right)^2}\\\\\\\to \boxed{\sf =\frac{\sec ^2\left(x\right)\left(1+x^2\right)-2x\tan \left(x\right)}{\left(1+x^2\right)^2}}$