Finding the partial derivatives of $h(x)=\int_{0}^{\|x\|} f(t)\, dt$ Find the partial derivatives of $$h(x_1,\dots,x_n)=\int_{0}^{\|x\|} f(t) dt$$ where $\|x\|$ is the Euclidean norm of $x=(x_1,\dots,x_n)$ and $f$ is some continuous function.
I'm sorry but I'm really not too sure how to approach this. Any help would be great!
(This is not homework, I'm preparing for an exam.)
 A: Note that if $$F(s) = \int^{b(s)}_{a(s)}f(t)dt,$$
then
$$F'(s) = b'(s)f(b(s))-a'(s)f(a(s)).$$
In particular if $$b(s) = \|(x_1,\ldots,x_{i-1},s,x_{i+1},\ldots,x_n)\|_2 \quad \text{ and } \quad a(s)=0,$$
then
$$b'(s) = \frac{s}{\|(x_1,\ldots,x_{i-1},s,x_{i+1},\ldots,x_n)\|_2} \quad \text{ and } \quad a'(s)=0,$$
It follows that 
$$\frac{\partial}{\partial x_i} h(x_1,\ldots,x_{i-1},s,x_{i+1},\ldots,x_n) = F'(s) =  \frac{s f(\|(x_1,\ldots,x_{i-1},s,x_{i+1},\ldots,x_n)\|_2)}{\|(x_1,\ldots,x_{i-1},s,x_{i+1},\ldots,x_n)\|_2}.$$
which can be written 
$$\frac{\partial}{\partial x_i} h(x) = \frac{x_i}{\|x\|_2}f(\|x\|_2),$$
with $s=x_i$.
A: $ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $ 
Hint: (see here for more information or the good answer provided by @Surb)
$$\small{\pd{}{x_1} \int^{\sqrt{x_1^2 +x_2^2 +\ldots + x_n^2}}_0f(s)\, \mathrm{d}s = f\left(\sqrt{x_1^2 +x_2^2 +\ldots + x_n^2}\right) \frac{x_1}{\sqrt{x_1^2 +x_2^2 +\ldots + x_n^2}} = \frac{f(\|x\|)}{\| x \|}x_1,}$$
can you generalize the result for any $x_i$? Sure you can.
Cheers!
A: Note that
\begin{equation*}
\frac{d}{dx_{j}}=\frac{d\parallel x\parallel }{dx_{j}}\frac{d}{d\parallel
x\parallel }=\frac{x_{j}}{\parallel x\parallel }\frac{d}{d\parallel
x\parallel }
\end{equation*}
