# why $\frac{d}{dy}$ can pass through integral w.r.t. $x$?

When I calculate integration of multivariables, many books use the following step without proofing. I want to know that why it is true: $$\frac{d}{dy}\left[\int^a_b f(x,y)dx\right]_{y=k}=\int^a_b \frac{\partial}{\partial y} \left[f(x,y)\right]_{y=k}dx$$

I also wonder that whether it is true when the integral or differentiation become indefinite. Which is : $$\frac{d}{dy}\int f(x,y)dx=\int\frac{\partial}{\partial y} f(x,y)dx$$ $$\frac{d}{dy}\int^a_b f(x,y)dx=\int^a_b \frac{\partial}{\partial y} f(x,y)dx$$ $$\frac{d}{dy}\left[\int f(x,y)dx\right]_{y=k}=\int \frac{\partial}{\partial y} \left[f(x,y)\right]_{y=k}dx$$

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This is an inverse version of the formula $\partial^2 u/\partial x \partial y = \partial^2 u/\partial y \partial x$. – GEdgar Jul 14 '13 at 13:39

As for your other question, the general indefinite integral of $f(x,y)$ is $\int f(x,y)\mathrm{d} x=\int_{x_0}^xf(t,y) \mathrm{d}t+C$ where $x_0,C$ are constants. Applying the rule here gives the required result (again if the requirements of the theorem are met).
If $f$ and $\frac{\partial f}{\partial y}$ are continuous in a closed rectangle, they are uniformly continuous there. consider the quotient $\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}$. Using the continuity hypothesis you can show that it tends to $\int f_y \mathrm d x$. The full proof can be found on the wiki page. You can find a counterexample here hss.caltech.edu/~kcb/Notes/LeibnizRule.pdf – user1337 Jul 14 '13 at 10:45