Differentiating Under the Integral Proof There are many variations of "differentiating under the integral sign" theorem; here is one:
If $U$ is an open subset of $\mathbb{R}^n$ and $f:U \times [a,b] \rightarrow \mathbb{R}$ is continuous with continuous partial derivatives $\partial_1 f, \dots \partial_n f$ then the function
$$
\phi(x) = \int^b_a f(x,t)dt
$$
is continuously differentiable and
$$
\partial_i \phi (x) = \int^b_a \partial_i f(x,t)dt
$$
Can anyone suggest a textbook that provides a proof of this version of the theorem?
 A: Isn't the proof sort of "follow your nose"? Let $\Delta x$ be nonzero, consider
$$ \phi(x+\Delta x)-\phi(x) = \int^{b}_{a}f(x+\Delta x,t)-f(x,t)\,\mathrm{d}t$$
Then construct the quotient
$$ \frac{\phi(x+\Delta x)-\phi(x)}{\Delta x} = \frac{\int^{b}_{a}f(x+\Delta x,t)-f(x,t)\,\mathrm{d}t}{\Delta x} $$
But because we do not integrate over $x$, we treat $x$ like a constant. So we can rewrite the integral as
$$ \frac{\phi(x+\Delta x)-\phi(x)}{\Delta x} = \int^{b}_{a}\frac{f(x+\Delta x,t)-f(x,t)}{\Delta x}\,\mathrm{d}t $$
Taking the limit as $\Delta x\to0$ gives us
$$ \frac{\mathrm{d}\phi(x)}{\mathrm{d} x} = \int^{b}_{a}\frac{\partial f(x,t)}{\partial x}\,\mathrm{d}t $$
precisely as desired? [Edit: We can take the limit under the integral sign, as Giuseppe Negro points out, if the function $f(x,t)$ is continuously differentiable in $x$.]
Addendum: Why, oh why, do we need $f(x,t)$ to be continuously differentiable in $x$?
Why can we take this limit? Well, there's a number of different arguments.
One is the Dominated convergence theorem, which states if we have a sequence of functions $f_{n}(t)\to F(t)$ which is "dominated" by some function $g(t)$, meaning
$$ |f_{n}(t)|\leq g(t)\quad\mbox{for any }t $$
then we have
$$ \lim_{n\to\infty}\int|f_{n}(t)-F(t)|\,\mathrm{d}t=0 $$
which implies
$$ \lim_{n\to\infty}\int f_{n}(t)\,\mathrm{d}t=\int F(t)\,\mathrm{d}t. $$
Take $F(t)=\partial f(x,t)/\partial x$ and $f_{n}(t)$ to be
$$ f_{n}(t) = \frac{f(x + \varepsilon_{n},t)-f(x,t)}{\varepsilon_{n}} $$
using any sequence $\varepsilon_{n}\to 0$. 
Addendum 2: A second different way begins with the observation
$$ \int^{b}_{a}\int^{x}_{0}\frac{\partial f(y,t)}{\partial y}\,\mathrm{d}y\,\mathrm{d}t = \phi(x)-\phi(0)$$
by the fundamental theorem of calculus. Fubini's theorem lets us switch the order of integration
$$ \int^{x}_{0}\int^{b}_{a}\frac{\partial f(y,t)}{\partial y}\,\mathrm{d}t\,\mathrm{d}y = \phi(x)-\phi(0)$$
Then we can use Leibniz's rule differentiating both sides with respect to $x$. This gives us the desired result
$$ \int^{b}_{a}\frac{\partial f(x,t)}{\partial x}\,\mathrm{d}t = \phi'(x).$$
Recall Leibniz's rule states if $G(x) = \int^{x}_{0}g(y)\,\mathrm{d}y$ then
$$ G'(x) = g(x). $$
We can prove this quickly by
$$ \frac{G(x+\Delta x)-G(x)}{\Delta x} = \frac{1}{\Delta x}\int^{x+\Delta x}_{x} g(y)\,\mathrm{d}y$$
and taking $\Delta x$ to be "sufficiently small", we can approximate the Riemann sum as
$$ \int^{x+\Delta x}_{x} g(y)\,\mathrm{d}y\approx g(c)\Delta x$$
where $x\leq c\leq x+\Delta x$. Plugging this back in gives us
$$ \frac{G(x+\Delta x)-G(x)}{\Delta x} = \frac{1}{\Delta x}\left(g(c)\Delta x\right) = g(c)$$
Taking $\Delta x\to 0$ gives us $c\to x$, and
$$ G'(x) = g(x)$$
as desired.
