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So we were asked that if $F(x,y,z)=0$,

show $\frac{\partial z}{\partial y}|_x \frac{\partial y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$.

Now I had to go about and show that $\frac{\partial z}{\partial y}|_x = -\frac{dz}{dy}$ etc, then substitute the partial derivates for their derivative equivalents and then I had the same thing as before but with standard derivatives instead of partials, and 3 negative signs. I then simplified the expression as if they were a bunch of fractions and the 3 negatives gave me a negative overall so I ended up with -1 as the answer.

But why can one not just treat partial derivates like fractions the same way you can for standard derivatives.

(I study Physics so I've been used to treating derivates as fractions for a long time)

Vishal Jain
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    A more fruitful question to ask instead of “why can one not?” is perhaps “why should one be able to do that to begin with?” There's simply no reason why that would work, if you think about what partial derivatives mean. By the way, your proof sketch seems a bit weird (the part about $-dz/dy$). You ought to have expressions involving derivatives of $F$ instead; see this answer, for example. – Hans Lundmark Oct 06 '19 at 19:24
  • I tried reading that answer but for some reason I cant see why you can say this " three quantities x, y and z constrained by a relation F(x,y,z)=0, so that one can view either one of them as a function of the two others: x=x(y,z) or y=y(x,z) or z=z(x,y)." – Vishal Jain Oct 06 '19 at 20:12
  • The explanation is in the following sentence: "this is the implicit function theorem". – Hans Lundmark Oct 06 '19 at 20:30
  • Ah yea after a bit more thinking it makes sense, just felt weird at first. – Vishal Jain Oct 08 '19 at 08:02

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On your surface, the differentials of $x$,$y$ and $z$ are related by $$ \frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=0. $$ If you consider $z$ as an (implicitly defined) function of $x,y$, to find $\frac{\partial z}{\partial y}$ you make $dx=0$ and solve to get $$ \frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}. $$ You can do the same for the other two. Multiply your results together and things cancel nicely to give $-1$.

GReyes
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