How to find the charge density given by a Tricky Potential?

[flux!]

Homework Statement

I am trying to solve Problem 2.45 in Electrodynamics by Griffiths, however, my answers were different from those in the book, I am suspect I got a missing step but I could not find it, so here is the Given Problem

Find the charge density \rho given by a potential V=A\frac{e^{-\lambda r}}{r} where A and \lambda are constant.

Homework Equations

E=-\nabla V \nabla\cdot E = \frac{\rho}{\varepsilon_0}

The Attempt at a Solution

So what I did is, first, find for the E-field using: E=-\nabla V, so E=-\hat{r}\frac{\partial }{\partial r} A\frac{e^{-\lambda r}}{r} by employing chain rule, the E-field is now given by E= A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r}.

Now, the charge density \rho can be obtained thru \rho=\varepsilon_0\nabla\cdot E, substituting the E-field to the equation, we will get \rho=\varepsilon_0\nabla\cdot \left [A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r} \right ], evaluating again using chain rule, we will get \rho=\varepsilon_0 \left [A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r} +\frac{1}{r}\left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A\lambda e^{-\lambda r} \right ) + A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r^2} +\frac{1}{r^2} \left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A e^{-\lambda r} \right ) \right ], this will eventually become a bit more messy but the charge density will become \rho=A \varepsilon_0 e^{-\lambda r} [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2}{r} ] however, this a bit different from Griffiths' answer of \rho=A \varepsilon_0 [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2 e^{-\lambda r}}{r}]

where did I went wrong?

 

[vela]

Use the delta function property $f(x)\delta(x) = f(0)\delta(x)$.

 

[flux!]

OMG! That was just the best!