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I have $z=f(x,y)$ known and I'd like to express $\partial z / \partial x$ as a function of $\partial y / \partial x$ and $\partial y / \partial z$. I know the solution is $$\frac{\partial z}{\partial x} = - \frac{\partial y / \partial x}{\partial y / \partial z}$$ But I can't get it myself. When I apply the chain rule I get $$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}$$ which clearly doesn't lead to the good result. Can you help me ?

rhorho
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  • $\partial x / \partial x = 1$ so I get $$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x}1 + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}$$ and $$0=\frac{\partial z}{\partial y} \frac{\partial y}{\partial x}$$ which doesn't look good – rhorho May 23 '15 at 09:47
  • Is $y$ a function of $x$? If not, you'll have $\frac{\partial y}{\partial x} = 0$. – MathNewbie May 23 '15 at 10:03

2 Answers2

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I'll restate my case after some rechecking my books.

We have that $$\frac{\partial z}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}$$ Notice however that $\frac{\partial x}{\partial x}=1$ and $\frac{\partial y}{\partial x}=0$ as I assume $y\neq f(x)$. this means you get it to be correct. $$\frac{\partial z}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y} 0=\frac{\partial z}{\partial x}$$ This of course mean that your initial formula has the same issue $$-\frac{\partial y/\partial x}{\partial y/\partial z}=\frac{0}{\partial y/\partial z}=0$$

If however $y=f(x)$ then what you really have isn't $g(x,y)$ but $g(x,f(g))=h(x)$ which is one variable and we can skip a lot of work.

Zelos Malum
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  • $y$ is function of $x$ and $z$ so we can't make the statement $ y \neq f(x)$ nor $y$ is only function of $x$ I think – rhorho May 23 '15 at 10:01
  • You had $z=g(x,y)$ at which we have two options, either x and y are independed or they are depended, that is $x=f(y)$ for some function. – Zelos Malum May 23 '15 at 10:02
  • The chain rule is primarely for when they depend on other things, such as $x=f_x(s,t)$ and $y=f_y(s,t)$ and the likes, when the variables are independed of each other, or one depended on the other, it is not applicable in the same way anymore and it collapses into being ordinary chainrule of one variable or just normal derivation without chainrule. – Zelos Malum May 23 '15 at 10:08
  • Ok thanks but I don't see how to use 1 variable chainrule as I don't know the relation linking y to x. It doesn't appear in the final formula I want to recover (the one I wrote above) – rhorho May 23 '15 at 10:12
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I'll give another more detailed one again, the chain rule only applies to functions "in" functions. That is $f=g\circ h$ and the likes, in our case we have $z=f(x,y)$ but is it a function in a function? It can be viewed as it only so much that $(x,y)=i(s,t)=(s,t)$ and then we have that $g=f\circ i(s,t)$ but the derivate for $\partial i/\partial s = \partial i/\partial t = 1$ so it doesn't add much at all. So chain rule isn't applicable here to begin with which is why you get like $\partial y/\partial x=0$ and the likes. We can on the other hand have for example $y=g(x)$ but then our $f(x,y)=f(x,g(x)=h(x)$ which we can redefine to be a function of one variable and just do then like normal.

But to answer your question on how to get it $$\frac{\partial z}{\partial x}=\frac{\partial z/\partial y}{\partial x/\partial y}=\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}=\frac{\partial y/\partial x}{\partial y/\partial z}$$ I think the negative sign is a typo on theo ther hand. What I did was simply expand the fraction with $(\partial y)^{-1}$ and then normal rules of arithmetic which is applicable.

Zelos Malum
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