I was asked to do this problem $\displaystyle \frac{d}{dx} |3x-x^2|=\frac{d}{dx} y$
I used the fact that $\displaystyle \frac{d}{dx} |x|= \frac{|x|}{x}$ so,
$\displaystyle \frac{|3x-x^2|(3-2x)}{3x-x^2}= \frac{dy}{dx}$
but my study mate did this; he squared both sides to get rid of the absolute value.
$9x^2-6x^3+x^4=y^2$ then he takes the derivative implicitly.
$\displaystyle \frac{9x-9x^2+2x^3}{y}=\frac{dy}{dx}$
I have tried some values and these both work reasonably, but I cant understand if they are the same?