# Help understanding a partial derivative $\frac{\partial\pi}{\partial C}$

This is one of my first problems involving partial derivatives and I don't understand how the partial works out to be $\frac{1}{d}$ given that $\pi=\frac{C}{d}$

I have to find:

$\frac{\partial\pi}{\partial C}$ and $\frac{\partial\pi}{\partial d}$

And I don't see how they'd turn out to be anything other than $0$ since, as I understand it, $C$ and $d$ are treated as constants depending on which derivative I'm taking.

## 2 Answers

As you see $\pi$ is a function of two variables $C$ and $d$. In common notation $$\pi(C,d)=\frac{C}{d}$$ which is like $f(x,y)$ in most problems. Assume now that someone asks you "to take the derivative of the function $\pi$". You ask naturally: "Ok, but with respect to which variable, $C$ or $d$?". "Aha, I should have specified that", he says and continues, "with respect let's say to the variable $C$". Then writting $$(\pi)'=\left(\frac{C}{d}\right)$$ does not indicate whether you take the derivative with repsect to $C$ or with respect to $d$. This problem is overcomed when we use the notation $$\frac{\partial \pi}{\partial C}$$ Now, think that $C$ is the variable $x$ you are used to and take the derivative of the function with respect to $C$, treating the other variable, i.e. $d$ as constant! That is $$\frac{\partial \pi}{\partial C}=\frac{1}{d}\cdot\frac{\partial}{\partial C}(C)=\frac{1}{d}\cdot1=\frac{1}{d}$$ Similarly $$\frac{\partial \pi}{\partial d}=C\frac{\partial}{\partial d}\left(\frac{1}{d}\right)=C\cdot\frac{-1}{d^2}=-\frac{C}{d^2}$$

• Perfect explanation! It makes so much since now. Thank you! – hax0r_n_code Mar 15 '14 at 22:27
• @inquisitor You are welcome, mate! – Jimmy R. Mar 15 '14 at 22:28

d(pi)/dC = 1/d because C/d = C*(1/d) and d is fixed. Likewise, d(pi)/dd = -C/d^2 because C is fixed.