Automatic way to derive $∫_0^x f(x,t) \,\mathrm{d} t$ Correct me if I'm wrong, but
\begin{align}
F(x) = ∫_a^b f(x,t) \,\mathrm{d} t &\implies F'(x) = ∫_a^b \frac{∂f}{∂t} \,\mathrm{d} t, \qquad \text{and} \\
G(x) = ∫_a^x f(t) \,\mathrm{d} t &\implies G'(x) = f(x).
\end{align}
But is there any general “formula” to derive this function
$$
F(x) = ∫_a^x f(x,t) \,\mathrm{d} t,
$$
or, in a more general way
$$
\frac{\mathrm{d}}{\mathrm{d}x} \biggl( ∫_{a(x)}^{b(x)} f(x,t) \, \mathrm{d}t \biggr).
$$
 A: Yes, apply Leibniz' integral rule (a.k.a. differentiation under the integral sign), 

$$\frac{\mathrm{d}}{\mathrm{d}x}\int_{a(x)} ^{b(x)} f(x,t)\,\mathrm{d}t
=f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)} ^{b(x)}\frac{\partial f}{\partial x}(x,t)\,\mathrm{d}t$$

which in your case gives
$$F'(x)=f(x,x)+\int_a ^x \frac{\partial f}{\partial x}(x,t)\,\mathrm{d}t.$$
Edit: As asked by a user, the precise conditions for Leibniz' integral rule to hold for $t\in[t_1,t_2]$ are


*

*$a,b$ are $C^1$ on $[t_1,t_2]$ 

*$f$ is $C^1$ on $\{(x,t): t\in [t_1,t_2], x\in[a(t),b(t)]\}$


(see, e.g., Kaplan: Advanced Calculus, Chpt. 4.9)
A: This is almost identitical to the question I asked and answered myself here:
Exact smoothness condition necessary for differentiation under integration sign to hold.
You need some additional smoothness condition, e.g. the partial derivative $\partial_xf(x,t)$ to be continuous. The wiki link is actually pretty dreadful as the smooth conditions are not clearly stated. I wonder if someone can give an authentic source.
