Differentiating under integral with bounded derivatives I remember there is a trick, where for certain conditions of a function $f:\mathbb{R}\rightarrow\mathbb{R}$, we can say that
$$\frac{d}{dx}\int_X f(x)d\mu(x) = \int_X\frac{d}{dx}f(x)d\mu(x).$$
The condition has something to do with the derivative of $f$ being bounded, and the proof of the statement goes by dominated convergence theorem. What is the exact statement and proof?
 A: You are probably thinking of this:
$$\frac{d}{dx}\int_\Omega f(x,y)\,d\mu(y) = \int_\Omega\frac{\partial f}{\partial x}(x,y)\,d\mu(y)$$
provided there is an integrable function $g$ with $$\left|\frac{\partial f}{\partial x}(x,y)\right|\le g(y).$$
The proof is by just writing up the derivatives as limits and applying DCT.
More precisely, you need to have $f\colon\mathbb{R}\times\Omega\to\mathbb{R}$ where $\mu$ is a measure on $\Omega$, $f$ needs to be differentiable at $x$ for almost all $y\in\Omega$ and measurable wrt $y$, and the above inequality needs to hold for almost every $y$ and all $x$.
Here is the gist of the proof:
$$\begin{aligned}
\frac{d}{dx}\int_X f(x,y)\,d\mu(y)
  &=\lim_{h\to0}\int_\Omega\frac{f(x+h,y)-f(x,y)}{h}\,d\mu(y) \\
  &=\int_\Omega\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}\,d\mu(y) \\
  &=\int_\Omega\frac{\partial f}{\partial x}(x,y)\,d\mu(y).
\end{aligned}$$
In the first line is the definition of the derivative except I skipped a step: You should have difference of two integrals and then divided by $h$, but I just joined the two integrals into one. In the second line I put the integral on the inside, which must be justified; and finally, the definition of the partial derivative.
The justification for taking the limit inside the integral is this: The mean value theorem from calculus gives
$$\frac{f(x+h,y)-f(x,y)}{h}=\frac{\partial f}{\partial x}(x+\theta h,y)$$
for some $\theta\in(0,1)$ (depending on $x$ and $y$), so the absolute value of the fraction is at most $g(y)$ by the assumption; so the DCT applies.
