TITLE: Derivative of a squared definite integral
QUESTION [2 upvotes]: Differentiation of
$\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t$ is done by Leibniz's integral rule:
$$\frac{\text{d}}{\text{d}x} \left (\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t \right )= f\big(x,b(x)\big)\cdot \frac{\text{d}}{\text{d}x} b(x) - f\big(x,a(x)\big)\cdot \frac{\text{d}}{\text{d}x} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,\text{d}t,$$
if $-\infty<a(x),b(x)<\infty$.
Can we say anything in general about the derivative of the powers of the $\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t$ in a similar fashion? So for example
$$\frac{\text{d}}{\text{d}x} \left(\left(\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t\right)^2\right)=~?$$

REPLY [3 votes]: Hint: According to Chain rule,
$$\frac{d}{dx} (y^n) = ny^{n-1}\frac{d}{dx}(y).$$

REPLY [3 votes]: By chain rule
$$\frac{\text{d}}{\text{d}x} \left(\left(\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t\right)^2\right)=2\left(\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t\right)\frac{\text{d}}{\text{d}x} \left(\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t\right)$$
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