# Thread: Gradient vector help

1. ## Gradient vector help

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

a) $\displaystyle \triangledown\\r=\frac{\vec{r}}{r}$, where $\displaystyle \vec{r}=<x,y>$

b) $\displaystyle \triangledown\\f(r)=$$\displaystyle f'(r)\triangledown\\r=$$\displaystyle \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 $\displaystyle f(r)$ 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 $\displaystyle r$ and then evaluate that as my function for $\displaystyle r$?

Any suggestions?

Thanks,

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

2. chain rule

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

$\displaystyle = f'(r) \nabla r$