1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\frac{\partial \pi}{\partial d}=C\frac{\partial}{\partial d}\left(\frac{1}{d}\right)=C\cdot\frac{-1}{d^2}=-\frac{C}{d^2}$$

$\endgroup$
  • $\begingroup$ Perfect explanation! It makes so much since now. Thank you! $\endgroup$ – hax0r_n_code Mar 15 '14 at 22:27
  • $\begingroup$ @inquisitor You are welcome, mate! $\endgroup$ – Jimmy R. Mar 15 '14 at 22:28
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.