Let $x = h(y, z), y = g(x, z), x = h(y, z)$ to calculate partial derivatives?

Problem: A $\mathbb{R}^{3}$ surface is defined by $F(x,\ y,\ z)=k$, where $k$ is a constant. Prove $\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1$.

I don't see the first step to this problem, which is to let $x = h(y, z), y = g(x, z), x = h(y, z)$. Can someone clarify this please?

Then when I differentiate $F(x,\ y,\ z)=c$, how can I figure out which of these functions, in terms of the other two, to use?

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Perhaps you mean $z=m(x,y)$? Because your third condition is a repetition of the first. – Mark Fantini May 30 '14 at 19:54

Pauline, this is a great and very important question. It might be quite tricky for those who haven't seen this before. However, if you very carefully think it through and master this problem, there's a good chance you'll really understand common uses of partial derivatives well.

In multivariate calculus, when dealing with partial derivatives, it's absolutely crucial that you always know what function you are talking about and where (at which point) you are evaluating partial derivatives. A big part of knowing what function you're dealing with is knowing what what coordinates, i.e., what "free parameters", are used to define that function. Unfortunately, the usual notations often obscure these crucial details. Furthermore, in this case, it's important you remember you are on a 2-dimensional surface, a 2-manifold. The formula in this exercise would not be true (or even make sense!) in other contexts.

On the 2-dimensional surface in $\mathbb{R}^3$ in this problem, you can compute any one of $x$, $y$, and $z$ in terms of the other two. Make sure you very clearly understand that. A little more precisely, this means there are functions $f$, $g$, and $h$ such that if $(x,y,z)$ is a point on the surface, then \begin{eqnarray} x &=& f(y,z) \\ y &=& g(x,z) \\ z &=& h(y,z). \end{eqnarray} You can flip back and forth between which two of $x$, $y$, and $z$ you think of as freely varying coordinates.

I'll given a very simple example. Consider the equation $$x^3 + y + z = 1.$$ This defines a curved 2-dimensional surface in $\mathbb{R}^3$. The points $(1,0,0)$ and $(-2,1,8)$ are on the surface, but $(3,3,3)$ isn't. Now if you know the $y$ and $z$ coordinates of a point on the surface, you can compute the $x$ coordinate: $$x = \sqrt[3]{1 - y - z}.$$ Thus, the function $f$ for this example is $(y,z) \mapsto \sqrt[3]{1 - y - z}$. Similarly, we have \begin{eqnarray} g(x,z) &=& 1 - x^3 - z \\ h(x,y) &=& 1 - x^3 - y. \end{eqnarray} for this example.

Now look at what you are trying to prove: $$\frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1.$$ The notation makes this formula look very simple and somehow almost obvious. But the simple notation is very misleading. There's a lot more complexity here than meets the eye.

Start out with the first factor on the left, $\frac{\partial x}{\partial y}$. A clearer way to write this quantity is $$\left(\frac{\partial f(y,z)}{\partial y}\right)_z.$$ Again, this is nothing more than a different notation for $\frac{\partial x}{\partial y}$. There's no new mathematical content here at all, but it makes what's happening clearer.

The denominator of the partial derivative ($y$) and the subscript ($z$) tell you your coordinates. The coordinates for the this partial derivative are $(y,z)$. The numerator ($x$) represents the function you are differentiating. And this function has to be a function of the coordinates $(y,z)$. So in the expression $\frac{\partial x}{\partial y}$, $x$ is not a coordinate, it's a function of $(y,z)$. Here, $x$ means the "value of the $x$ coordinate of the point on the surface with given $y$ and $z$ coordinates". In other words, in the expression $\frac{\partial x}{\partial y}$, $x$ actually represents the function $(y,z)\mapsto f(y,z)$.

The subscript $z$ and the denominator $y$ also tell you the direction in your coordinate space along which the derivative is taken. Here, we hold $z$ constant and vary $y$. This is important. Always be clear on what's being held constant (more generally, the direction along which you are differentiating).

So you are really trying to prove the following formula: $$\left(\frac{\partial f(y,z)}{\partial y}\right)_z \left(\frac{\partial g(x,z)}{\partial z}\right)_x \left(\frac{\partial h(x,y)}{\partial x}\right)_y = -1.$$ Stare at that for a second. There are three different coordinate spaces $(y,z)$, $(x,z)$ and $(x,y)$. Each of $x$,$y$, and $z$ is a coordinate in two of these partial derivatives and a function of the other coordinates in the third.

Now, to prove it! Assume $F(x,y,z) = c$ implicitly defines our 2-manifold. Consider $(y,z)$ as our coordinates, so $x=f(y,z)$ for any point on the manifold. Thus, $$F(f(y,z), y, z) = c.$$ The left hand side is a function of only $y$ and $z$. Now compute the partial derivative with respect to the coordinate $y$ holding $z$ constant. We use the chain rule to obtain \begin{eqnarray} \left(\frac{\partial F(f(y,z), y, z)}{\partial y} \right)_z &=& \frac{\partial F}{\partial x} \frac{\partial f}{\partial y} + \frac{\partial F}{\partial y}. \end{eqnarray} Here on the right hand side, for brevity, I've gone back to the usual very terse notation that suppresses the exact function parameters and partial derivative subscripts.

Now because $F$ is constant on the surface, this partial derivative must vanish: $$\frac{\partial F}{\partial x} \frac{\partial f}{\partial y} + \frac{\partial F}{\partial y} = 0,$$ and $$(*)\ \ \ \frac{\partial F}{\partial x} \frac{\partial f}{\partial y} = - \frac{\partial F}{\partial y}.$$

Now repeat this procedure with coordinates $x,z$ and $y=g(x,z)$. Compute the partial derivative of $F$ with respect to $z$ (holding $x$ constant). The computation is exactly the same as above, just with the indices swapped. We obtain $$(*)\ \ \ \frac{\partial F}{\partial y} \frac{\partial g}{\partial z} = - \frac{\partial F}{\partial z}.$$

Repeat the procedure one last time with coordinates $x$ and $y$ and $z=h(x,y)$. Differentiate with respect to $x$ holding $y$ constant. One obtains $$(*)\ \ \ \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = - \frac{\partial F}{\partial x}.$$

Now just multiply the left corresponding sides of the three equations marked $(*)$ and you get $$\frac{\partial F}{\partial x} \frac{\partial F}{\partial y} \frac{\partial F}{\partial z} = - \frac{\partial F}{\partial x} \frac{\partial F}{\partial y} \frac{\partial F}{\partial z} \frac{\partial f}{\partial y} \frac{\partial g}{\partial z} \frac{\partial h}{\partial x}.$$ As long as $\frac{\partial F}{\partial x} \frac{\partial F}{\partial y} \frac{\partial F}{\partial z} \ne 0$, which is one of those pesky simplifying assumptions, we have $$\frac{\partial f}{\partial y} \frac{\partial g}{\partial z} \frac{\partial h}{\partial x} = -1.$$

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