**Proof plan.** We write out the difference quotient $\frac{I(x + \delta x) - I(x)}{\delta x}$, split the resulting [integrals](/page/Integral) into three pieces (the shared region $[a(x), b(x)]$, the upper [boundary](/page/Boundary) strip near $b$, and the lower boundary strip near $a$), and take the [limit](/page/Limit) $\delta x \to 0$ using the [fundamental theorem of calculus](/theorems/632) for the boundary contributions and the definition of the partial [derivative](/page/Derivative) for the integrand contribution. **Step 1: Form the difference quotient.** \begin{align*} \frac{I(x+\delta x) - I(x)}{\delta x} = \frac{1}{\delta x}\left[\int_{a(x+\delta x)}^{b(x+\delta x)} f(t; x+\delta x) \, dt - \int_{a(x)}^{b(x)} f(t; x) \, dt\right]. \end{align*} **Step 2: Split into three contributions.** Add and subtract $\int_{a(x)}^{b(x)} f(t; x + \delta x) \, dt$: \begin{align*} \frac{I(x+\delta x) - I(x)}{\delta x} &= \frac{1}{\delta x} \int_{a(x)}^{b(x)} \bigl[f(t; x+\delta x) - f(t; x)\bigr] \, dt \\ &\quad + \frac{1}{\delta x} \int_{b(x)}^{b(x+\delta x)} f(t; x+\delta x) \, dt \\ &\quad - \frac{1}{\delta x} \int_{a(x)}^{a(x+\delta x)} f(t; x+\delta x) \, dt. \end{align*} **Step 3: Take the limit of each piece.** As $\delta x \to 0$: The first integral converges to $\int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(t; x) \, dt$ by passing the limit inside the integral (justified by the [continuity](/page/Continuity) of $\partial f / \partial x$). For the second integral, by the [mean value theorem](/theorems/186) for integrals: \begin{align*} \frac{1}{\delta x} \int_{b(x)}^{b(x+\delta x)} f(t; x+\delta x) \, dt \to f(b(x); x) \frac{db}{dx}. \end{align*} Similarly, the third integral converges to $f(a(x); x) \frac{da}{dx}$. **Step 4: Combine.** \begin{align*} \frac{dI}{dx} = \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(t; x) \, dt + f(b(x); x)\frac{db}{dx} - f(a(x); x)\frac{da}{dx}. \end{align*}