Problem Let $u = f(r)$ and $r = |\mathbf{x}| =(x_1^2+\dots+x_n^2)^{1/2}$. Show that $\sum_{1}^{n} (\dfrac{\partial{u}}{\partial{x_j}}) ^ {2} = [f'(r)]^2$
I expanded the LHS to be $$ \sum_{1}^{n} (\frac{\partial u}{\partial x_j})^2 = (\frac{\partial{f}}{\partial{r_1}} \frac{\partial{r_1}}{\partial{x_1}} + \dots + \frac{\partial{f}}{\partial{r_n}} \frac{\partial{r_n}}{\partial{x_1}})^2 + (\frac{\partial{f}}{\partial{r_1}} \frac{\partial{r_1}}{\partial{x_2}} + \dots + \frac{\partial{f}}{\partial{r_n}} \frac{\partial{r_n}}{\partial{x_2}})^2 + \dots (\frac{\partial{f}}{\partial{r_1}} \frac{\partial{r_1}}{\partial{x_n}} + \dots + \frac{\partial{f}}{\partial{r_n}} \frac{\partial{r_n}}{\partial{x_n}})^2 $$
Is this correct? I have no idea how to evaluate the RHS. Any help is appreciated!