# Prove the correctness of Leibniz's rule for differentiation under the integral sign on $0<x<1$

Good day all, here is a question I have concerning Leibniz's rule for differentiation under the integral sign. Let \begin{align} G(x)=\int^{\beta(x)}_{\alpha(x)}g(x,t)dt\end{align} (a.) Place conditions on $\alpha,\;\beta$ and $g$ so that $G'(x)$ exists for $0<x<1.$

(b.) Give the formula for computing $G'(x)$ and prove its correctness.

(a.) $\alpha(x)$ and $\beta(x)$ must be continuous and must have continuous derivatives for $0<x<1.$ Also, $g(x,t)$ must be a function such that $g_x(x,t)$ are continuous in $t$ and $x$ in some region of $(x,t)-$plane including $\alpha(x)\leq x \leq \beta(x),\;0<x<1.$
Hint: Assume there exists a function $H$ so that $\frac{\partial H}{\partial t} = g$. Then $\int_I g(x,t) dt = H(x,t)]_I$ then expand on the interval H, then differentiate the two terms you'll get.