Prove fromulae for $\text{ trace}(\nabla^2\varphi )$ and $\det (\nabla^2 \varphi)$ for radial symmetric function Show that when a function $\varphi$ is radial (i.e. is only a function of $r = \sqrt{x^2 + y^2}$ ) and $\frac{d\varphi}{dr}$ = $\varphi^{'}$ and  $\frac{d^2\varphi}{dr^2}$ = $\varphi^{''}$, then:
$$\text{trace}(\nabla^2\varphi(x,y) = \varphi^{''}(r) + \frac{\varphi^{'}(r)}{r}     $$
and
$$  \det(\nabla^2\varphi(x,y)) = \frac{\varphi^{'}(r)\varphi^{''}(r)}{r}  .$$
We just learned that $\nabla^2$ is the Hessian and the professor said this problem is easy, however, I'm not sure where to start.
 A: I'd do half of the question (for $\text{tr} \nabla^2 \varphi$) and you can try to finish the second half: Note 
$$\begin{split} 
\frac{\partial ^2 \varphi}{\partial x^2} &= \frac{\partial }{\partial x} \left(\frac{\partial \varphi}{\partial x}\right) \\
&= \frac{\partial }{\partial x} \left( \frac{\partial \varphi}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial \varphi}{\partial \theta} \frac{\partial \theta}{\partial x}\right) \\
&= \frac{\partial }{\partial x} \left( \frac{\partial \varphi}{\partial r} \frac{\partial r}{\partial x} \right) 
\end{split}$$
since $\varphi$ is independent of $\theta$. Applying product rule and chain rule again, 
$$\begin{split} 
\frac{\partial }{\partial x} \left( \frac{\partial \varphi}{\partial r} \frac{\partial r}{\partial x} \right) &= \frac{\partial }{\partial x} \left( \frac{\partial \varphi}{\partial r}\right)\frac{\partial r}{\partial x}+   \frac{\partial \varphi}{\partial r}\frac{\partial^2 r }{\partial x^2}\\
&=\frac{\partial^2 \varphi}{\partial r^2} \left(\frac{\partial r}{\partial x}\right)^2+ \frac{\partial \varphi}{\partial r}\frac{\partial^2 r }{\partial x^2}
\end{split}$$
Since 
$$\frac{\partial r}{\partial x} = \frac xr,\ \ \ \frac{\partial ^2 r}{\partial x^2} = \frac{y^2}{r^3},$$
we conclude 
$$\frac{\partial^2 \varphi}{\partial x^2} = \frac{x^2}{r^2} \varphi '' (r)  + \frac{y^2}{r^3}\varphi'(r).$$
Similarly, 
$$\frac{\partial^2 \varphi}{\partial y^2} = \frac{y^2}{r^2} \varphi '' (r)  + \frac{x^2}{r^3}\varphi'(r),$$
thus 
$$ \text{tr} \nabla^2 \varphi =\frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}= \varphi'' + \frac 1r \varphi'. $$
A: Hint: With the chain rule, we can write
$$
\varphi'(r) = \varphi_x x_r
+ \varphi_y y_r =
\varphi_x \cos \theta+ \varphi_y \sin \theta=\\
\varphi_x \frac x r + \varphi_y \frac y r
$$
Where subscripts denote partial derivatives. You can do something similar with $\varphi''(r)$.
