# How to find the partial derivative of this function?

Lets say I have a function:$$\nu=\frac{RT}{P}+B_{p}(T)RT$$ and I am trying to find $\left(\frac{\partial \nu}{\partial T}\right)_{P}$. I understanding that the partial derivative of the first term is just $\frac{R}{P}$ but the second term has two terms that depend on T. Do I use the product rule like regular derivatives? If so, I think the answer would look something like this:

$$\left(\frac{\partial \nu}{\partial T}\right)_{P}=\frac{R}{P}+RT \frac{\partial B_{P}(T)}{\partial T}+RB_{P}(T)$$

Am I correct in saying this or does the product rule not apply to partial derivatives like I was thinking.

• This is the right way to do this, yes. I'm not sure about your notation. Why $\left(\frac{\partial v}{\partial T}\right)_P$ and not just $\frac{\partial v}{\partial T}$ in the left hand side ? – Pixel Dec 8 '13 at 8:15
• Happy to see some one here playing with the mathematics in equations of state ! – Claude Leibovici Dec 8 '13 at 8:46
• @pbs, the notation is saying the change in $\nu$ with $T$ while keeping $P$ constant. That is the notation my professor has given us and also how it is shown in the Thermodynamics book. I was just trying to be consistent – Greg Harrington Dec 8 '13 at 19:18
• @GregHarrington I see. In that case wouldn't $\left(\frac{\partial v}{\partial T}\right)_{P,R}$ be more appropriate... – Pixel Dec 9 '13 at 12:57