Show that if $\chi(r)=rR(r)$ then $\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=r\frac{d^2 \chi(r)}{dr^2}$ My Attempt:
I first applied the product rule to 
$$\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}\tag{1}\label1$$ to obtain 
$$2r\frac{dR(r)}{dr}+r^2\frac{d^2R(r)}{dr^2}\tag{2}\label2,$$ and then wrote out the chain rule for 
$$\frac{dR(r)}{dr}$$ as 
$$\frac{dR(r)}{dr}=\frac{dR(r)}{d \chi(r)}\frac{d\chi(r)}{dr}\tag{3}\label3.$$
Then from $\chi(r)=rR(r)$ and using the product rule again 
$$\frac{d\chi(r)}{dr}=R(r)+r\frac{d^2R(r)}{dr^2}\tag{4}\label4,$$ and since $R(r)=\dfrac{\chi(r)}{r},$ then
$$\frac{dR(r)}{d \chi(r)}=\frac{1}{r}\tag{5}\label5.$$
Substituting $\eqref4$ and $\eqref5$ into $\eqref3$, I find that 
$$\frac{dR(r)}{dr}=\frac{1}{r}\left(R(r)+r\frac{d^2R(r)}{dr^2}\right)=\frac{1}{r}R(r)+\frac{d^2R(r)}{dr^2}\tag{6}\label6$$
Substituting $\eqref6$ into $\eqref2$ yields 
$$\begin{align}\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}&=2r\frac{dR(r)}{dr}+r^2\frac{d^2R(r)}{dr^2}\\&=2r\left(\frac{1}{r}R(r)+\frac{d^2R(r)}{dr^2}\right)+r^2\frac{d^2R(r)}{dr^2}\\&=2R(r)+(2r+r^2)\frac{d^2R(r)}{dr^2}.\end{align}$$ 

As you can see the expression I desire for $$\fbox{$\color{blue}{\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)}$}$$ is getting every-more complicated and I haven't even computed the second derivative of $R(r)$ yet. I fear that I am either making a mistake or missing a much simpler approach. In my notes it said that

It is straightforward to show that $$\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)=r\frac{d^2 \chi(r)}{dr^2}.$$

So this implies that I am missing something very simple.
If someone would care to point out my errors and/or give me any hints/tips on how to reach the desired result I would be most thankful. Regards.
 A: I write $f$ instead of $R$ and $g$ instead of $\chi$. Then it is to show:
$\frac{d}{dr}(r^2f'(r))=rg''(r)$
From $g(r)=rf(r)$ we get $g'(r)=f(r)+rf'(r)$, hence $r^2f'(r)=rg'(r)-rf(r)=rg'(r)-g(r)$,
thus
$\frac{d}{dr}(r^2f'(r))=g'(r)+rg''(r)-g'(r)=rg''(r)$, as desired.
A: Well, we have that:
$$\mathcal{Z}=\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}\left(\text{r}\right)\right)\right)=\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\cdot\text{R}'\left(\text{r}\right)\right)=\text{r}^2\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}'\left(\text{r}\right)\right)+\text{R}'\left(\text{r}\right)\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\right)\tag1$$
Using the product rule.
So, we get:
$$\mathcal{Z}=\text{r}^2\cdot\text{R}''\left(\text{r}\right)+\text{R}'\left(\text{r}\right)\cdot2\text{r}=\text{r}^2\text{R}''\left(\text{r}\right)+2\text{r}\text{R}'\left(\text{r}\right)\tag2$$
And we also have:
$$\text{r}\cdot\chi\space''\left(\text{r}\right)=\text{r}\cdot\frac{\text{d}^2}{\text{d}\text{r}^2}\left(\text{r}\cdot\text{R}\left(\text{r}\right)\right)=\text{r}\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{R}\left(\text{r}\right)\right)+\text{R}\left(\text{r}\right)\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\right)\right)\tag3$$
