Let $f: \mathbb R^n \setminus \{0\} \to \mathbb R^n \setminus \{0\}$ be such that $f$ only depends on the distance from the origin, that is, $f=f(r)$ where $r = \sqrt{\sum_{i=1}^n x_i^2}$.
I am trying to calculate $\Delta f$.
Could someone please tell me if this is right?
My work:
First, I calculated the derivative of $r$ with resepct to one variable:
$$ {\partial r \over \partial x_i} = {1\over 2}\cdot \left( \sum_{i=1}^n x_i^2\right)^{-{1\over 2}}\cdot 2x_i = x_i r$$
Then I calculated the first partial derivative:
$$ {\partial f \over \partial x_i}f(r) = {\partial r\over \partial x_i}{\partial f\over \partial r} = x_i r f_r$$
and then the second partial derivative:
$$ {\partial^2 f \over \partial x_i^2}f(r) = {\partial \over \partial x_i}x_i r f_r = {\partial \over \partial x_i}(x_i r)\cdot f_r + x_i r \cdot {\partial \over \partial x_i} f_r = r + x_i^2 rf_r + (x_ir)^2f_{rr}$$
Then
$$ \Delta f = \sum_{i=1}^n {\partial^2 f\over \partial x_i^2} = nr + r^3f_r + r^4f_{rr}$$
There is an error in your first calculation: $$ \frac{\partial r}{\partial x_i} = \frac{1}{2} \Bigl( \sum_{i=1}^n x_i^2\Bigr)^{-{1\over 2}}\cdot 2x_i = \frac{x_i }{r}. $$