Answer
$f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$
Work Step by Step
\[
\text{Step 1: Apply the Fundamental Theorem}\\
\text{ of Calculus (FTC)} \\
\int_{g(x)}^{h(x)} f(t)\,dt = F(h(x)) - F(g(x))
\]
\[
\text{Step 2: Differentiate both sides}\\
\text{ with respect to } x \\
\frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(t)\,dt \right) = \frac{d}{dx} [F(h(x)) - F(g(x))]
\]
\[
\text{Step 3: Apply the chain rule }\\
\text{to each term} \\
f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)
\]
\[
\text{Final Result:} \\
\frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(t)\,dt \right) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)
\]
