Question

Derivada de
$y=2xcotgx+x^2tgx$

Answer 1

$\\ y = 2xCotgx +x^2tgx \\ \\ y' = (2x)'*Cotgx +2x*cotgx'+(x^2)'tgx + x^2tgx' \\ \\ y' = 2cotgx +2x*(-cossec^2x)+2xtgx +x^2*Sec^2x \\ \\ y' = 2cotgx -2xcossec^2x+2xtgx+x^2Sec^2x$

Answer 2

$\sf \displaystyle y=2x\:cotg\:x+x^2tgx\\\\\\\frac{d}{dx}\left(2xcy\tan \left(x\right)+x^2\tan \left(x\right)\right)\\\\\\{Tratar}\:c,\:y\:{como\:constante}\\\\\\=\frac{d}{dx}\left(2xcy\tan \left(x\right)\right)+\frac{d}{dx}\left(x^2\tan \left(x\right)\right)\\\\\\\to \boxed{\sf =2cy\left(\tan \left(x\right)+x\sec ^2\left(x\right)\right)+2x\tan \left(x\right)+\sec ^2\left(x\right)x^2}$