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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 ^_^

kram1032
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Cecile
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1 Answers1

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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 $$\frac{dz}{dy}\frac{dy}{dx}\frac{dx}{dz}=\bigg(-\frac{\partial F/\partial y}{\partial F/\partial z}\bigg)\bigg(-\frac{\partial F/\partial x}{\partial F/\partial y}\bigg)\bigg(-\frac{\partial F/\partial z}{\partial F/\partial x}\bigg)=-1$$

AnilB
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  • 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. – user599310 May 20 '20 at 16:30
  • Rather late to answer @user599310 's question, but the (missing) context of this question is that we are discussing defining $x, y, z$ as functions of each other by $F(x,y,z)=0$ on a connected neighborhood of a point where the partial derivatives of $F$ with respect to the three variables are non-zero. By the Implicit function theorem, each of the variables can be expressed as a unique function of the others in this neighborhood. For your example $F(x,y,z) = x+y^2 - z$, the partial derivative when $y = 0$ is $0$, so the neighborhood cannot include points with $y = 0$. – Paul Sinclair Nov 25 '23 at 16:34
  • This means that $y$ is either everywhere positive or everywhere negative within that neighborhood, allowing us to express either $y = \sqrt{z-x}$ or $y=-\sqrt{z-x}$. – Paul Sinclair Nov 25 '23 at 16:37