Finding the partial derivatives of $h(x)=\int_{0}^{\|x\|} f(t)\, dt$

Question

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.)

Answer 1

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 $s=x_i$.

Answer 2

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*}

Answer 3

$ \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!