[proofplan] We derive the density of $Y = h(X)$ by differentiating the cdf. We compute $F_Y(y) = \mathbb{P}(h(X) \le y)$, handling the two cases (increasing and decreasing $h$) separately, and differentiate using the chain rule. The absolute value of $(h^{-1})'(y)$ unifies both cases. [/proofplan] [step:Compute the cdf of $Y = h(X)$ in both the increasing and decreasing cases] **Case 1: $h$ is strictly increasing.** Since $h$ is strictly increasing with inverse $h^{-1}$, the inequality $h(X) \le y$ is equivalent to $X \le h^{-1}(y)$ for $y \in \operatorname{Im}(h)$. Therefore \begin{align*} F_Y(y) = \mathbb{P}(Y \le y) = \mathbb{P}(X \le h^{-1}(y)) = F_X(h^{-1}(y)). \end{align*} **Case 2: $h$ is strictly decreasing.** The inequality $h(X) \le y$ reverses to $X \ge h^{-1}(y)$, giving \begin{align*} F_Y(y) = \mathbb{P}(X \ge h^{-1}(y)) = 1 - F_X(h^{-1}(y)). \end{align*} [/step] [step:Differentiate the cdf using the chain rule to obtain the pdf] **Case 1 ($h$ increasing).** Since $h^{-1}$ is differentiable (by the inverse function theorem, as $h'$ is non-vanishing), we differentiate $F_Y(y) = F_X(h^{-1}(y))$ using the chain rule: \begin{align*} f_Y(y) = F_Y'(y) = f_X(h^{-1}(y)) \cdot \frac{d}{dy} h^{-1}(y). \end{align*} When $h$ is increasing, $h^{-1}$ is also increasing, so $\frac{d}{dy} h^{-1}(y) > 0$ and the absolute value is unnecessary. **Case 2 ($h$ decreasing).** Differentiating $F_Y(y) = 1 - F_X(h^{-1}(y))$, \begin{align*} f_Y(y) = -f_X(h^{-1}(y)) \cdot \frac{d}{dy} h^{-1}(y). \end{align*} When $h$ is decreasing, $h^{-1}$ is also decreasing, so $\frac{d}{dy} h^{-1}(y) < 0$ and the negative sign makes $f_Y(y) > 0$. [/step] [step:Unify the two cases using the absolute value] In both cases, the density of $Y$ is \begin{align*} f_Y(y) = f_X(h^{-1}(y)) \left|\frac{d}{dy} h^{-1}(y)\right|. \end{align*} When $h$ is increasing, the absolute value has no effect (the derivative is positive). When $h$ is decreasing, the absolute value absorbs the negative sign from the chain rule and the negative sign from differentiating $1 - F_X$. [/step]