The scalar field $f$ depends only on $r=|\underline{\mathbf{r}}|$ which is the position vector in $\mathbf{R}^3$ and I need to calculate the quantity, $$\nabla \cdot \nabla f$$ i went about calculating $\nabla f$ and I got this,
$$(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x},\frac{\partial f}{\partial r}\frac{\partial r}{\partial y},\frac{\partial f}{\partial r}\frac{\partial r}{\partial z})$$ $$ = (f'\frac{x}{r},f'\frac{y}{r},f'\frac{z}{r})$$ $$= f' \frac{\underline{r}}{r} \quad (*) $$ where $f'=\frac{\partial f}{\partial r}$ and i got the $\frac{x}{r}$ from differentiating the position vector.
I'm fairly certain that's okay but my problem comes with taking the divergence so i have that, $$\nabla \cdot \nabla f = (\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z})\cdot\frac{1}{r}(f'x,f'y,f'z)$$ i pull out the $\frac{1}{r}$ from the dot product and use the chain rule.$$\frac{1}{r}(\frac{\partial (f'x)}{\partial x}+\frac{\partial (f'y)}{\partial y}+\frac{\partial (f'z)}{\partial z}) $$ which when using the chain rule gives $$\frac{1}{r}(3f'+x\frac{\partial f'}{\partial x}+y\frac{\partial f'}{\partial y}+\frac{\partial f'}{\partial z})$$ Now heres where im unsure i say this is equivalent to, $$\frac{1}{r}(3f'+\underline{r}\cdot (\nabla f'))$$ we use $(*)$ to get $$\frac{1}{r}(3f'+\underline{r}\cdot(f''\frac{\underline{r}}{r}))$$ $$= \frac{1}{r}(3f'+(\underline{r}\cdot\underline{r})(f''\frac{1}{r}))$$ using the fact that $x \cdot x = |x|^2$ i find that $$\nabla \cdot \nabla f =3\frac{f'}{r}+f''$$
This is my first time really using differential operators in this way so any help checking is much appreciated.
That is not allowed! $r = r(x, y, z)$, so you can't just ignore. You have
\begin{eqnarray} \nabla \cdot \nabla f(r) &=& \nabla \cdot \left(f' \frac{{\bf r}}{r} \right) \\ &=& \frac{\partial}{\partial x}\left(\frac{xf'}{r} \right) + \frac{\partial}{\partial y}\left(\frac{y f'}{r} \right) + \frac{\partial}{\partial z}\left(\frac{z f'}{r} \right) \\ &=& \left(\frac{f'}{r} + \frac{x f''}{r}\frac{x}{r} - \frac{x^2f'}{r^3}\right) + \left(\frac{f'}{r} + \frac{y f''}{r}\frac{y}{r} - \frac{y^2f'}{r^3}\right) + \left(\frac{f'}{r} + \frac{z f''}{r}\frac{z}{r} - \frac{z^2f'}{r^3}\right) \\ &=& \frac{3f'}{r} + \frac{f''}{r^2}(x^2 + y^2 + z^2) - \frac{f'}{r^3}(x^2 + y^2 + z^2) \\ &=& 3\frac{f'}{r} + r^2\frac{f''}{r^2} - r^2\frac{f'}{r^3} \\ &=& 2\frac{f'}{r} + f'' \end{eqnarray}