The task is as follows:

Given: $$F(x,y,z) = 0$$ Goal: Show $\frac{\partial z}{\partial y}|_x \frac{\partial y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$

Here is my work so far:

(1) Differentiate with respect to y, I get:

$0 + F_2 + F_3 \frac{\partial z}{\partial y} = 0$

So $ F_3 \frac{\partial z}{\partial y} = - F_2$

(2) Differentiate with respect to x, I get:

$F_1 + F_2 \frac{\partial y}{\partial x} + 0 = 0$

So $ F_2 \frac{\partial y}{\partial x} = - F_1$

(3) Differentiate with respect to z, I get:

$F_1 \frac{\partial x}{\partial z} + 0 + F_3 = 0$

So $ F_1 \frac{\partial x}{\partial z} = - F_3$

(4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z} = -1$, so when evaluated with x, z, y respectively, conclusion is still true

My questions are:

(1) Is my proof correct?

(2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.

I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing. Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too. Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y. But is my thought ok?

Would someone please help me on this question? Thank you very much ^_^


1 Answer 1


A more simple way is the total differential $$dF=\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0$$ If it says "evaluated at $x$" it means that $x$ is fixed and $dx=0$ $$\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dz}{dy}=-\frac{\partial F/\partial y}{\partial F/\partial z}$$ and similarly $$dy=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dx}{dz}=-\frac{\partial F/\partial z}{\partial F/\partial x}$$ $$dz=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy=0\Rightarrow \frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}$$ And it follows $$\frac{dz}{dy}\frac{dy}{dx}\frac{dx}{dz}=\bigg(-\frac{\partial F/\partial y}{\partial F/\partial z}\bigg)\bigg(-\frac{\partial F/\partial x}{\partial F/\partial y}\bigg)\bigg(-\frac{\partial F/\partial z}{\partial F/\partial x}\bigg)=-1$$

  • $\begingroup$ If we let $z=y^2+x$ how we can treat $y$ as a function of $x$. It is like saying that $y^2=x−z$ is a function. $\endgroup$
    – user599310
    May 20, 2020 at 16:30

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.