Q: let $r=\sqrt{x^2+y^2}$. Show that...

a) $\triangledown\\r=\frac{\vec{r}}{r}$, where $\vec{r}=$

b) $\triangledown\\f(r)=$ $f'(r)\triangledown\\r=$ $\big({\frac{f'(r)}{r}})\vec{r}$

I got the first part just fine, but I am stuck on part b. I'm not sure what $f(r)$ really is. If I let the derivative equal 1, then everything falls into place and I get equivalent answers, but I doubt thats how I am supposed to do it. Do I solve for $r$ and then evaluate that as my function for $r$?

Any suggestions?

Thanks,

Oh, and how do I make my parenthesis proportional to what’s inside them?! Its angering me....

2. chain rule

$\nabla f(r) = \nabla f(\sqrt{x^{2}+y^{2}}) = \Big (f'(\sqrt{x^{2}+y^{2}} * \frac {x}{\sqrt{x^{2}+y^{2}}}$ $, f'(\sqrt{x^{2}+y^{2}}) * \frac{y}{ \sqrt{x^{2}+y^{2}}} \Big)$

$= f'(r) \nabla r$