How to find the derivative of $F(t) = \int_0^t f(t, x) \, dx$? Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continuous function. Define $F: \mathbb{R} \to \mathbb{R}$ by,
$$
F(t) = \int_0^t f(t, x) \, dx
$$
Then, I'm not sure how to get $F'$. If, there are functions $g$ and $h$ such that $f(t, x) = g(t)h(x)$, then of course we have
$$
F(t) = \int_0^t f(t, x) \, dx = \int_0^t g(t)h(x) \, dx = g(t)\int_0^t h(x) \, dx
$$
which can then be differentiated using the product rule. But apart from this special case, I don't know how to get $F'$.
 A: Let $g(u, t) = \int_0^u f(t,x)dx$.  Let  $u = t$ and apply the chain rule.
A: Use the Leibniz Rule as demonstrated here:
$$
{d\over dt}\int_0^t f(x,t)\,dx=\int_{0}^{t}{\partial f\over \partial t}\,dx+f(t,t).
$$
A: $$
{\rm F}\left(t\right)
=
{\rm sgn}\left(t\right)\int_{0}^{\infty}\Theta\left(\left\vert t\right\vert - x\right)
{\rm f}\left(t,x\,{\rm sgn}\left(t\right)\right)\,{\rm d}x\,,
\qquad\quad
\phi\left(t,x\right) \equiv {\partial{\rm f}\left(t,x\right) \over \partial t} 
$$
\begin{eqnarray*}
{\rm F}'\left(t\right)
& = &
\int_{0}^{\infty}\left\lbrack%
2\delta\left(t\right)\
\Theta\left(\left\vert t\right\vert - x\right)\
{\rm f}\left(t,x\,{\rm sgn}\left(t\right)\right)
+
{\rm sgn}\left(t\right)\
\delta\left(\left\vert t\right\vert - x\right)\ {\rm sgn}\left(t\right)\
{\rm f}\left(t,x\,{\rm sgn}\left(t\right)\right)
\right.
\\&&\phantom{\int_{0}^{\infty}\left\lbrack\right.}
+
\\&&
\left.
\phantom{\int_{0}^{\infty}\left\lbrack\right.}
\left.{\rm sgn}\left(t\right)\ \Theta\left(\left\vert t\right\vert - x\right)\
\phi\left(t,x'\right)\right\vert_{x'\ =\ x\,{\rm sgn}\left(t\right)}
\right\rbrack
\,{\rm d}x
\end{eqnarray*}
$$
{\rm F}'\left(t\right)
=
{\rm f}\left(t,t\right)
+
\int_{0}^{t}{\partial{\rm f}\left(t,x\right) \over \partial t}\,{\rm d}x
$$
