Originally Posted by

**Scott H** Here, the Chain Rule states

$\displaystyle \frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}.$

We have, by calculation

$\displaystyle \frac{df}{dr} = -\frac{1}{r^2}\;\;\;\;\;\;\;\;\;\;\frac{\partial r}{\partial x} = 2x.

$

Thus, we have

$\displaystyle \frac{\partial f}{\partial x}=\left(-\frac{1}{r^2}\right)\cdot(2x)=-\frac{2x}{r^2}.$

I'm not sure what you mean by $\displaystyle \mbox{grad}\,\Delta f$, but for $\displaystyle \nabla f$ I calculated something that looked quite different from the answer given in the problem.