About chain rule

Question

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 ?

Answer 1

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.

Answer 2

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.