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I'm having trouble with the following:

The equations $z = g(x,v)$ and $ y = f(x,v)$ can be thought of as defining $z$ as a function of $x$ and $y$, that is: $z = \phi(x,y)$. Show that: $$\frac{\partial \phi}{\partial y} = \left. \frac{\partial g}{\partial v} \middle/ \frac{\partial f}{\partial v} \right. $$

If I write out a small change in $g$ as $\delta g$ = $\delta \phi$ = $\frac{\partial \phi}{\partial y}\delta y + \frac{\partial \phi}{\partial x}\delta x$ and then divide by $\delta v$ and make the $\delta$s infinitesimal then I get $$\frac{\partial g}{\partial v} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial v}$$ and since $y = f$ this is nearly what I need.

How do I show that $\frac{\partial \phi}{\partial x}\frac{\partial x}{\partial v}$ = 0 so that I get to the answer?

Poo2uhaha
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  • what do you mean by $\frac{\partial \phi}{\partial y} = \frac{\partial g}{\partial v} / \frac{\partial f}{\partial v}$? $\phi$ is a function from $\mathbb R ^2 \to \mathbb R ^2$ so its partial derivative for one variable will still have to coordinates. – Targon Dec 24 '20 at 12:03
  • I think ϕ is a function from $\mathbb{R}^2$ -> $\mathbb{R}$ – Poo2uhaha Dec 24 '20 at 12:10
  • yeah sorry I misread – Targon Dec 24 '20 at 12:19
  • I don't want to go into infinitesimals because I can't. But $\frac {\partial x}{\partial v} = 0$ because $x$ doesn't depend on $v$. So if it's correct until that point, you're done. – Targon Dec 24 '20 at 12:29
  • I was thinking I might be able to use that, but in the case where v is a function of x, this method wouldn't cover that - and I think it might be supposed to work for all cases. But yeah, it doesn't seem like a massive issue; thanks for your help. – Poo2uhaha Dec 24 '20 at 12:35

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