Let $f $ be a real valued function on $\mathbb R^2$. Suppose that $f (x,y) $ is integrable as a function if $y $ for each $x $. Suppose that $\partial f /\partial x $ is defined at all $(x,y) \in \mathbb R^2$. Finally, suppose that there is an integrable function $g (y) $ on $\mathbb R $ such that $|\partial f /\partial x| \leq g (y) $ at all points in $\mathbb R^2$. Show that $$\frac {d}{dx} \int _{\mathbb R}f (x,y)dx = \int_{\mathbb R} \frac {\partial f}{\partial x} dy $$ at all $x \in \mathbb R $.
I think that the question meant to say $\frac {d}{dx} \int _{\mathbb R}f (x,y)dy$ instead of $dx $. Correct me if I am wrong.
I am not sure what to do here. Here is my attempt so far. Fix $x $. The question hints strongly at the dominated convergence theorem. Hence, letting $p_n(y) := \frac {f (x+1/n, y) - f (x,y)}{1/n} $, we see that $p_n \to \frac {\partial f}{\partial x}$ . If we can only show that $|p_n (y)| \leq g (y) $ then we'd be able to apply the dominated convergence theorem but I am not sure how to do that.
Firstly, you are right - it should be $dy$ inside the integral.
Secondly, using the assumption $|\partial_x f(x,y)|\leq g(y)$ for all $(x,y)\in\mathbb{R}^2$, and the Mean value theorem, we get for some $\theta \in [0,1]$, $$ \left\vert \frac{f(x+h,y)-f(x,y)}{(x+h)-x} \right\vert = \left\vert\partial_x f(x+\theta h,y)\right\vert \leq g(y) \quad\Rightarrow \quad \left\vert \frac{f(x+h,y)-f(x,y)}{h} \right\vert \leq g(y). $$ The Dominated convergence theorem solves the rest, as you have already mentioned.