Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the variable $x,y,z$ possibly depending on each other i.e. on a smooth manifold. Using the theory of differential forms I can derive $\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial y}{\partial z}\right)_x=-1$ by \begin{align}dx\wedge dz=\left(\frac{\partial x}{\partial y}\right)_z dy\wedge dz=\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial z}{\partial x}\right)_y dy\wedge dx =\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial y}{\partial z}\right)_x dz\wedge dx\end{align}

I don't understand the formal mathematical meaning of the subscripts. They are indicating which variable is held constant (why do I need this information?).

Now I should proof the following for the variables $x,y,z,w$:\begin{align} \left(\frac{\partial x}{\partial y}\right)_z=\left(\frac{\partial x}{\partial y}\right)_w+\left(\frac{\partial x}{\partial w}\right)_y\left(\frac{\partial w}{\partial y}\right)_z\end{align}

Not only that I don't know how to prove it, I don't even understand the different meaning of $\left(\frac{\partial x}{\partial y}\right)_z$ and $\left(\frac{\partial x}{\partial y}\right)_w$. Any advice regarding the concrete problem and some reading advice on the general topic are very welcome . I tried Schutz's geometrical methods of mathematical physics, but I'd like something with a focus on purer mathematics.

share|cite|improve this question
Regarding the subscripts, suppose that $x = y + z$. What is $\left( \frac{\partial x}{\partial y} \right)_z$? What is $\left( \frac{\partial x}{\partial y} \right)_{-y + z}$? – Qiaochu Yuan Oct 22 '12 at 20:45
Is $\left(\frac{\partial x}{\partial y}\right)_z=1$ and $\left(\frac{\partial x}{\partial y}\right)_{-y+z}=\left(\frac{\partial (2y-y+z)}{\partial y}\right)_{-y+z}=2$ correct? I think I see where this is going... – Julian Oct 22 '12 at 21:03
So in $\left(\frac{\partial x}{\partial y}\right)_z$ it is assumed that $w$ still can depend on $y$? I can write $x=x(y,z,w)$. What is $\left(\frac{\partial x}{\partial y}\right)_z$ then? – Julian Oct 22 '12 at 21:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.