Given pdf $f_{Y}(y)$ = $ay^2e^{-by^2}$, $y >=0$, find the pdf of the kinetic energy, $W = 0.5mY^2$ problem:
Suppose the velocity of a gas molecule of mass $m$ is a random variable with pdf $f_{Y}(y)$ = $ay^2e^{-by^2}$, $y >=0$, where $a$ and $b$ are positive constants depending on the gas. Find the pdf of the kinetic energy, $W = 0.5mY^2$, of such a molecule.
Attempt: By definition
\begin{align}
F_{W} &= P(W \le w) \\&= P( 0.5mY^2 \le w) \\&= P(Y^2 \le \frac{2w}m)
\\&=P(Y\le  \frac{\sqrt{2w}}m)\end{align}
Then $f_{W}(w) = d/dw [(F_{W}(w))] $.
Can someone please help me finish. I don't know how to continue. I am stuck in finding the derivative. Please I would really appreciate it if someone helps me. Thank you.

...Hence $$f_W(w)=\frac{\mathrm d}{\mathrm dw}\int_0^{\sqrt{2w}/m}f_Y(y)\mathrm dy.$$
What is the general formula for $$\frac{\mathrm d}{\mathrm dw}\int_{A(w)}^{B(w)}g(y)\mathrm dy,$$ already?
