# Help Needed: Partial Derivative Identity/Chain Rule

I'm taking a calculus class and I need help proving the following. I'm looking for a starting point and not the entire solution.

Consider three variables $x$, $y$, and $z$, satisfying a functional relation $f(x,y,z)=0$ for some function $f$. Let $w$ be another function of any two of $x$, $y$, or $z$. Show $\left(\frac{\partial x}{\partial y}\right)_{w}\left(\frac{\partial y}{\partial z}\right)_{w}=\left(\frac{\partial x}{\partial z}\right)_{w}$.

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@Max Do you mean Let $w$ be another function of any two of $x,y$ or $z$? – user38268 Feb 8 '12 at 3:16
@BenjaminLim Yes. Thanks. – Max Feb 8 '12 at 4:43
What does the subscript $w$ mean? – Dylan Moreland Feb 8 '12 at 4:49
@DylanMoreland I believe it means that $w$ is held constant, but honestly I'm pretty lost. – Max Feb 8 '12 at 5:00

You have three variables and one relation, so two variables can be chosen independently. Thus when writing a partial derivative with respect to one variable, it suffices to note the second variable that is being held constant, in this case $w$. See my answer here for a collection of answers in which I tried to resolve confusions about partial derivatives stemming mostly from our suboptimal notation for them.

In the present case, as in general, the key is to keep clear about what is being considered as a function of which variables, and explicit notation helps a lot in doing that, so you may want to write the right-hand side more explicitly as

$$\def\deriv#1#2{\frac{\partial #1}{\partial #2}} \deriv{x(w,z)}z\;.$$

The result then follows if you apply the general rule

$$\deriv{g(s,t)}t=\deriv{g(u,v)}u\deriv{u(s,t)}t+\deriv{g(u,v)}v\deriv{v(s,t)}t$$

with suitable $u,v$ and use

$$\deriv{w(w,z)}w=0\;.$$

The function $f$ doesn't enter into the calculation and is just there to ensure that only two and not three of the variables are independent.

There is also a conceptually simpler way to arrive at the result. With both $f$ and $w$ being held constant, we only have one degree of freedom left. Thus if we keep in mind that this only makes sense in the framework where $f$ and $w$ are considered fixed, we can consider $x$, $y$ and $z$ as ordinary functions of each other and simply write

$$\def\oderiv#1#2{\frac{\mathrm d#1}{\mathrm d#2}} \oderiv xy\oderiv yz=\oderiv xz\;,$$

which is just the chain rule for $x(y(z))$.

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