exchanging partial derivative and an integral

Question

Does $\frac{\partial }{\partial x}\int_{u(x)}^{v(x)}f(x,t)dt$ = $\int_{u(x)}^{v(x)}\frac{\partial }{\partial x}f(x,t)dt$ ??

If yes, how did we do that although there are 2 functions $u$ and $v$ in the integration limits?

Here is a screenshot from a book that made me confusion

Answer 1

Let $I(x,u,v) = \int_u^v f(x,t) dt$. Suppose $f$ is sufficiently smooth, for example $f$ and ${\partial f(x,t) \over \partial x}$ are continuous.

Then ${\partial I(x,u,v) \over \partial x} = \int_u^v {\partial f(x,t) \over \partial x} dt$, ${\partial I(x,u,v) \over \partial u} = -f(x,u)$ and ${\partial I(x,u,v) \over \partial v} = f(x,v)$.

Now let $\phi(x) = I(x,u(x),v(x))$, then the chain rule gives ${\partial \phi(x) \over \partial x} = {\partial I(x,u(x),v(x)) \over \partial x} + {\partial I(x,u(x),v(x)) \over \partial u}u'(x) + {\partial I(x,u(x),v(x)) \over \partial v}v'(x) $.

Expanding gives: ${\partial \phi(x) \over \partial x} = \int_{u(x)}^{v(x)} {\partial f(x,t) \over \partial x} dt + f(x,v(x)) v'(x) - f(x,u(x))u'(x)$.