Question

Derivada de
$y=-x^2ln x$

Answer 1

$\\ y = -x^2Lnx \\ \\ y' = -x^2'Lnx + -x^2Lnx' \\ \\ Y' = -2xLnx-x^2* \frac{1}{x} \\ \\ Y' = -2xLnx -x \\ \\ Y' = -x( 2Lnx +1) \\ \\ Y' = -x( Lnx^2+1) \\ \\ y' = -x(Lnx^2 + Lne) \\ \\ y' = -x( Lnx^2*e) \\ \\ y' = -xLn(ex^2)$

Answer 2

$\sf \displaystyle y=-x^2inx\\\\\\\frac{d}{dx}\left(-x^2inx\right)\\\\\\=-in\left(\frac{d}{dx}\left(x^2\right)x+\frac{d}{dx}\left(x\right)x^2\right)\\\\\\=-in\left(2xx+1\cdot \:x^2\right)\\\\\\\to \boxed{\sf =-3inx^2}$