Partial differential equation solution

Question

I'm trying to figure out how to go from: $${1 \over {r}}{({{∂P} \over{∂r}} + r {{∂^2 P} \over{∂r^2} })}= {1 \over {n}} {{∂P} \over {∂t}}$$

to:

$${1 \over {r}}{ ∂ \over{∂r} }{(r \times {{∂P} \over{∂r} })}= {1 \over {n}} {{∂P} \over {∂t}}$$

I'm rusty on my calculus and I'm trying to understand the deduction of the diffusivity equation.

Answer 1

\begin{eqnarray} { ∂ \over{∂r} }{(r \times {{∂P} \over{∂r} })} &=& \frac{\partial r}{\partial r}.\frac{\partial P}{\partial r}+r\frac{\partial }{\partial r}\frac{\partial P}{\partial r}=\frac{\partial P}{\partial r}+r\frac{\partial^2P}{\partial r^2} \end{eqnarray} so \begin{eqnarray} \frac1r{ ∂ \over{∂r} }{(r \times {{∂P} \over{∂r} })} &=& \frac1r\left(\frac{\partial P}{\partial r}+r\frac{\partial^2P}{\partial r^2}\right)={1 \over {n}} {{∂P} \over {∂t}} \end{eqnarray}