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What I've tried so far:

$$F(x,y,z(x,y)) = 0$$ $$\implies \frac{\partial F}{\partial x} = 0$$

By the chain rule:

$$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} + \frac{\partial F}{\partial x}\frac{\partial x}{\partial x} = 0$$ $$= \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} + \frac{\partial F}{\partial x} = 0$$

We know that $$\frac{\partial F}{\partial x} = 0$$ therefore

$$= \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} = 0$$

How do I proceed from here?

dfg
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  • How do you get $F_x=0?$ It is, $F_x+F_zz_x=0.$ – mfl Oct 19 '14 at 18:38
  • @mfl Well you know that $F = 0$, so $\frac{\partial F}{\partial x} = \frac{\partial 0}{\partial x} = 0$ – dfg Oct 19 '14 at 18:40
  • Take $F(x,y,z)=x+y+z.$ It is $F_x=1.$ – mfl Oct 19 '14 at 18:41
  • Note that in a similar way, it would be $F_y=0.$ So, is it $z_x=z_y=0?$ Absolutely not. – mfl Oct 19 '14 at 18:42
  • But isn't the partial derivative of a constant equal to zero? – dfg Oct 19 '14 at 18:42
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    Yes. But the constant is $0=F(x,y,z(x,y)).$ You have $x$ in the first and in the third entries. You have to derive both, using the chain rule. – mfl Oct 19 '14 at 18:43
  • Sorry, I don't follow. If you differentiate both sides with respect to $x$, the derivative of $0$ is zero no matter what. So why doesn't that imply that no matter what $F$ is, $F_x = 0$? – dfg Oct 19 '14 at 18:46
  • Consider, again, $0=x+y+z=F(x,y,z(x,y))=x+y+(-x-y),$ $z(x,y)=-x-y.$ You can't consider only the derivative on the first $x.$ $z$ depends on $x$ and so you have to consider its derivative with respect to $x.$ – mfl Oct 19 '14 at 18:49

2 Answers2

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You have

$$0=\frac{\partial 0}{\partial x}= \frac{\partial F(x,y,z(x,y))}{\partial x}=F_x+F_z z_x\implies z_x=-\frac{F_x}{F_z}.$$

In a completely similar way

$$0=\frac{\partial 0}{\partial y}= \frac{\partial F(x,y,z(x,y))}{\partial y}=F_x+F_z z_y\implies z_y=-\frac{F_y}{F_z}.$$

mfl
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  • Thank you. I understand what your doing, but this still confuses me: $F$ is just another way of writing $F(x,y,z(x,y)$ and $F_x$ is just another way of writing $\frac {\partial F}{\partial x}$. So isn't $\frac{\partial {F(x,y,z(x,y)}}{{\partial x}}$ just another way of writing $F_x$? I get the whole chain rule thing, but the notation is essentially implying that $F_x = F_x + F_zz_x$... – dfg Oct 19 '14 at 19:07
  • No. Forget for a moment that $F(x,y,z(x,y))=0.$ How would you compute the derivative with respect to $x$ of $F(x,y,x^2+y^2)?$ Consider, for example, $F(x,y,z)=x+z.$ Isn't it $\frac{\partial F(x,y,x^2+y^2)}{\partial x}=1+2x=F_x+F_zz_x?$ – mfl Oct 19 '14 at 19:10
  • I would use the chain rule to find $F_x$ and get something similar to what you did above. But exactly what part of my previous comment was incorrect? Does $F \neq F(x,y,z(x,y))$? – dfg Oct 19 '14 at 19:13
  • In my opinion, you are confusing $F(x,y,z)=0$ and $F(x,y,z(x,y))=0.$ In the first case, you are right, it is $F_x=0.$ We can fix $y,z$ on the left side and let $x$ to change. But it is always equal to zero. So, the derivative with respect to $x$ is zero. In the second case we have $F(x,y,z(x,y))=0.$ We can fix $y$ to be constant. But we can't fix $z$ to be constant because it depends on $x.$ This is where the part $F_zz_x$ enters the game. – mfl Oct 19 '14 at 19:18
  • I think I understand that - my problem is with the notation. How can you say $F(x,y,z(x,y)) = F$ but then say $F(x,y,z(x,y)_x \neq F_x$. Your answer helped me understand why they are not equal, but now I'm wondering about the notation. – dfg Oct 19 '14 at 19:23
  • I did some research and figured it out. $F(x,y,z(x,y))_x$ is the total derivative but $F_x$ is the partial derivative. – dfg Oct 19 '14 at 19:37
  • Yes, you are right. $F_x$ is the partial derivative when we consider $x,y,z$ as independent variables. But when we have some relation, say $z(x,y),$ then $F(x,y,z(x,y))_x$ means the derivative of $f(x)=F(x,y,z(x,y)).$ I think you have clarified you problem. Good luck. – mfl Oct 19 '14 at 19:41
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Note that the you can always fix $y$ and change only $x$, then you get the same setting as in the theorem. Of course, you can also fix $x$ and let $y$ vary. In fact, also when you have $f:\mathbb{R}^n\to\mathbb{R}$, you can always fix all the variables except for two, and use the theorem.

Amitai Yuval
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