# Vector Differentiation

• January 10th 2011, 05:38 AM
kjchauhan
Vector Differentiation

Prove that:

$\nabla^2 f(r) = f''(r) + \frac{2}{r}f'(r)$

Where $r^2=x^2+y^2+z^2$

• January 10th 2011, 06:30 AM
snowtea
Hint:

$\displaystyle \frac{\partial f(r)}{\partial x} = f'(r) \frac{\partial r}{\partial x} = f'(r)\frac{x}{r}$

$\displaystyle \frac{\partial^2 f(r)}{\partial^2 x} = \frac{\partial^2 f'(r)\frac{x}{r}}{\partial x} = f''(r)\frac{x^2}{r^2} + f'\frac{r^2 - x^2}{r^3}$.
• January 10th 2011, 06:50 AM
kjchauhan
Quote:

Originally Posted by snowtea
Hint:

$\displaystyle \frac{\partial f(r)}{\partial x} = f'(r) \frac{\partial r}{\partial x} = f'(r)\frac{x}{r}$

$\displaystyle \frac{\partial^2 f(r)}{\partial^2 x} = \frac{\partial^2 f'(r)\frac{x}{r}}{\partial x} = f''(r)\frac{x^2}{r} + f'\frac{r^2 - x^2}{r^3}$.

Thanks for Hint..
This is correct but I couldn't find the right hand side ..
According to your hint, the first term of the right will be $rf''(r)$..
So, is it correct??

Ok, well, got it..

Thank you very much..
• January 10th 2011, 06:52 AM
snowtea
Quote:

Originally Posted by kjchauhan
Thanks for Hint..
This is correct but I couldn't find the right hand side ..
According to your hint, the first term of the right will be $rf''(r)$..
So, is it correct??

I made a typo. It should be $\frac{x^2}{r^2}$. It is now fixed in the previous post.