Derivative with Respect to a Variable Times Constant? And Then Factoring-Out That Constant as a Fraction? I have the PDE $\dfrac{\partial{T}}{\partial{t}} = \alpha \dfrac{1}{r} \dfrac{\partial}{\partial{r}}\left( r \dfrac{\partial{T}}{\partial{r}} \right)$
I have done change of variables and now have $r^* = \dfrac{r}{a}$, $\dfrac{\partial{T}}{\partial{t}} = \dfrac{(T_i - T_w)}{t_0} \dfrac{\partial{u}}{\partial{t^*}}$, and $\dfrac{\partial{T}}{\partial{r}} = \dfrac{(T_i - T_w)}{a} \dfrac{\partial{u}}{\partial{r^*}}$.
I am told that the original PDE can be rewritten as $\dfrac{1}{t_0} \dfrac{\partial{u}}{\partial{t^*}} = \dfrac{\alpha}{a^2 r^*} \dfrac{\partial}{\partial{r^*}} \left( r^* \dfrac{\partial{u}}{\partial{r^*}} \right)$.
Doing the necessary substitutions, if my calculations are correct, we get 
$$\dfrac{(T_i - T_w)}{t_0} \dfrac{\partial{u}}{\partial{t^*}} = \dfrac{\alpha}{r^* a} \dfrac{\partial}{\partial{r}}\left( r^* a \left[ \dfrac{(T_i - T_w)}{a} \dfrac{ \partial{u} }{ \partial{r^*} } \right] \right)$$
And, if it is correct to do so (?), we substitute $r = r^* a$ into $\dfrac{ \partial }{ \partial{r}}$ to get  
$$\dfrac{(T_i - T_w)}{t_0} \dfrac{\partial{u}}{\partial{t^*}} = \dfrac{\alpha}{r^* a} \dfrac{\partial}{\partial{(r^* a)}}\left( r^* a \left[ \dfrac{(T_i - T_w)}{a} \dfrac{ \partial{u} }{ \partial{r^*} } \right] \right)$$
I have never seen such a thing, so I have no idea if this is even mathematically correct, but assuming it is, in order to get the rewritten PDE, we would have to factor out the $a$ to get
$$\dfrac{(T_i - T_w)}{t_0} \dfrac{\partial{u}}{\partial{t^*}} = \dfrac{\alpha}{a^2 r^*} \dfrac{\partial}{\partial{r^*}}\left( r^* a \left[ \dfrac{(T_i - T_w)}{a} \dfrac{ \partial{u} }{ \partial{r^*} } \right] \right)$$
And doing the necessary cancellations, we get 
$$\dfrac{1}{t_0} \dfrac{\partial{u}}{\partial{t^*}} = \dfrac{\alpha}{a^2 r^*} \dfrac{\partial}{\partial{r^*}}\left( r^* \dfrac{ \partial{u} }{ \partial{r^*} } \right)$$
But, as I said, I have never seen someone differentiating with respect to a variable multiplied by a constant, and then factoring out the constant as the fraction $\dfrac{1}{a}$ ($a$ in this case). Can someone please explain whether or not this is valid and why?
 A: It is perfectly possible, to show you why use the chain rule. Here's (perhaps) a simpler version, call
\begin{eqnarray}
T &=& (T_i - T_w)u  \\
r &=& a r^* \\
t &=& t_0 t^*
\end{eqnarray}
So that
$$
\frac{\partial }{\partial r} = \frac{{\rm d}r^*}{{\rm d}r}\frac{\partial }{\partial r^*} = \frac{{\rm d}(r/a)}{{\rm d}r}\frac{\partial}{\partial r^*} = \frac{1}{a}\frac{\partial}{\partial r^*}
$$
Similarly
$$
\frac{\partial}{\partial t} = \frac{1}{t_0}\frac{\partial}{\partial t^*}
$$
And your equation becomes
\begin{eqnarray}
\require{cancel}
\frac{1}{t_0}\frac{\partial}{\partial t^*}[\color{blue}{\cancel{(T_i - T_w)}}u] &=& \frac{\alpha}{a^2}\frac{1}{r^*}\frac{\partial }{\partial r^*}\left( \frac{\color{red}{\cancel{a}} r^*}{\color{red}{\cancel{a}}} \frac{\partial }{\partial r^*}[\color{blue}{\cancel{(T_i - T_w)}}u]\right) \\
\implies~~\frac{1}{t_0} \frac{\partial u}{\partial t^*} &=& \frac{\alpha}{a^2} \frac{1}{r^*}\frac{\partial}{\partial r^*}\left(r^* \frac{\partial u}{\partial r^*} \right)
\end{eqnarray}
