$\frac{d}{dx} \int_a^b f(x,t) dt =\int_a^b \frac{\partial}{\partial x}f(x,t)dt?$ I have studied Advanced Calculus by Fitzpatrick which discusses the so called the Second Fundamental Theorem (Differentiating Integrals) but it is when the upper or lower limits of integral is a function of $x$ and differentiation is on $x$. But differentiation of the type in the title is not discussed at all and is used in another textbook without a proof; so what is a clear proof for equation $$\dfrac{d}{dx} \int_a^b f(x,t) \ dt =\int_a^b \frac{\partial}{\partial x}f(x,t) \ dt?$$
 A: A proof would use some "limit theorem" to interchange integral and limit.  Probably in a course with only Riemann integration we would use uniform convergence for that.  But in a course with Lebesgue integration we could use a monotone convergence or dominated convergence argument.  Here we mark with $\overset{*}{=}$ where the limit theorem must be used.
\begin{align}
\frac{d}{dx} \int_a^b f(x,t)\;dt &=
\lim_{h\to 0}\frac{1}{h}\left(\int_a^b f(x+h,t)\;dt -  \int_a^b f(x,t)\;dt\right) 
\\ &=
\lim_{h\to 0}\int_a^b\frac{1}{h}\big(  f(x+h,t) - f(x,t)\big)\;dt
\\ &\overset{*}{=}
\int_a^b\lim_{h\to 0}\frac{1}{h}\big(  f(x+h,t) - f(x,t)\big)\;dt
\\ &=
\int_a^b \frac{\partial}{\partial x} f(x,t)\;dt
\end{align}
The remainder of the argument would use assumed properties of $f(x,t)$ in order to verify the hypotheses of the limit theorem you wish to use at $\overset{*}{=}$.
A: For a decently behaving $\frac{\partial}{\partial x}f(x,t)$ one has $f(x,t)=f(x_0,t)+\int_{x_0}^x\frac{\partial}{\partial y}f(y,t)\ dy\ $ and therefore $$\int_a^b f(x,t) \ dt=\int_a^b f(x_0,t) \ dt+\int_a^b\int_{x_0}^x\frac{\partial}{\partial y}f(y,t)\ dy\ dt=$$ $$=\int_a^b f(x_0,t) \ dt + \int_{x_0}^x\int_a^b\frac{\partial}{\partial y}f(y,t)\ dt\ dy$$
and $$\dfrac{d}{dx} \int_a^b f(x,t) \ dt=\dfrac{d}{dx}\int_{x_0}^x\int_a^b\frac{\partial}{\partial y}f(y,t)\ dt\ dy=\int_a^b \frac{\partial}{\partial y}f(y,t) \ dt$$
A: Recall Leibniz's integral rule
$$\frac{d}{dt}\int_{\phi(t)}^{\psi(t)} f(t,s) ds = \int_{\phi(t)}^{\psi(t)} \frac{d}{dt}f(t,s) ds+f(t,\psi(t))\frac{d}{dt}\psi(t) -f(t,\phi(t))\frac{d}{dt}\phi(t)$$
Notice that if ${\displaystyle \psi(t)}$and ${\displaystyle \phi(t)} $are constants rather than functions of ${\displaystyle t} $, we have a special case of Leibniz's rule:
$$\frac{d}{dt}\int_{a}^{b} f(t,s) ds = \int_{a}^{b} \frac{d}{dt}f(t,s) ds+f(t,b)\cdot 0 -f(t,a)\cdot0$$
$$\frac{d}{dt}\int_{a}^{b} f(t,s) ds = \int_{a}^{b} \frac{d}{dt}f(t,s) ds$$
The formula you are looking for is nothing but a special case of Leibniz's rule.
A: "Fundamental Theorem (Differentiating Integrals) but it is when the upper or lower limits of integral is a function of x and differentiation is on x"
might not count as a proof but just an observation I thought might be useful,
when you mentioned that the limits of the integral are a function of x, your limits in this case be taken as :
f1(x)=a and f2(x)=b
in which case the proof given in the book would suffice..
