Let $X$ be a continuous random variable with the pdf given by
$$ f(x) = \begin{cases} \frac{1}{\beta} e^{-\frac{x}{\beta}}, & x > 0 \\ 0, & \text{elsewhere} \end{cases} $$
Find the pdf of $Y = \sqrt{X}$.
My attempt:
$$F_Y(y) = P(Y\le y) = P(\sqrt{X}\le Y) = P(X\le Y^2) $$
$$ f_Y(y) = \begin{cases} \frac{2ye^{-\frac{y}\beta}}{\beta}, & \text{if } y >0 \\ 0, & \text{elsewhere} \end{cases} $$
Am I on the right path?