1
$\begingroup$

A portion of the heat diffusion equation for a 1-D solid is given as:

$$\frac{1}{r} \frac{\partial}{\partial r} \left(r \; k \frac{\partial T}{\partial r} \right)$$

Apparently this can be expanded into the following form using the Chain rule:

$$\frac{\partial}{\partial r} \left(k \frac{\partial T}{\partial r} \right) + \frac{1}{r} \left(k \frac{\partial T}{\partial r} \right)$$

However, the only way I can get this form is using the Product rule, not the Chain rule.

Is there a way to use the Chain rule or is the Product rule actually the method to use?

$\endgroup$
2
  • $\begingroup$ You are correct that is the product rule. $\endgroup$
    – user142299
    Commented Apr 14, 2014 at 2:19
  • $\begingroup$ @user142299 Are you sure it's the Product rule? I just finished reading a book on finite difference methods in heat transfer in which it states the chain rule in multiple locations in the book was used to get the above form. Maybe the book meant to say the product rule. I was just wondering if there was a way to use the chain rule too. $\endgroup$
    – wigging
    Commented Apr 14, 2014 at 2:25

1 Answer 1

1
$\begingroup$

This is definitely the product rule. Note that $$\frac{1}{r} \frac{\partial}{\partial r} \left(r \; k \frac{\partial T}{\partial r} \right)={1\over r}\left(\frac{\partial}{\partial r}(r)\cdot\left(k \frac{\partial T}{\partial r}\right)+\frac{\partial}{\partial r}\left(k \frac{\partial T}{\partial r}\right)(r)\right)=\frac{\partial}{\partial r} \left(k \frac{\partial T}{\partial r} \right) + \frac{1}{r} \left(k \frac{\partial T}{\partial r} \right)$$

$\endgroup$
4
  • $\begingroup$ Yes that looks like the approach I used. For the product rule, considering $u*v' + v*u'$, what did you make $u$ and $v$ equal to? $\endgroup$
    – wigging
    Commented Apr 14, 2014 at 2:40
  • $\begingroup$ $u=k\frac{\partial T}{\partial r}$, $v=r$. $\endgroup$
    – user142299
    Commented Apr 14, 2014 at 2:41
  • $\begingroup$ I ended up using $u = r k$ and $v = \frac{\partial T}{\partial r}$ which seemed to give the same answer. $\endgroup$
    – wigging
    Commented Apr 14, 2014 at 2:48
  • $\begingroup$ I think they used the $u,v$ that I showed above. That's how they got their formula. $\endgroup$
    – user142299
    Commented Apr 14, 2014 at 2:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .