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I have $h(x,y)=g(f(x,y))$

The relationship between $f$ and $g$ are expressed with $h$ $$f(x,y)=y+xh(x,y)$$

Prove that $$\frac{\partial f}{\partial x}(x,y) = h(x,y)\frac{\partial f}{\partial y}(x,y)$$

I tried from two sides

$$h(x,y)\frac{\partial f}{\partial y}(x,y) \\\ = h(x,y) \frac{\partial f}{\partial y}(y+xh) \\\ = h(x,y) (1+x)\frac{\partial h}{\partial y} \\\ = h+xh \frac{\partial f}{\partial y}(y+xh)$$

$$f(x,y)\frac{\partial f}{\partial x} \\\ =\frac{\partial }{\partial x}(y+xh) \\\ = \frac{\partial }{\partial x} xh \\\ = 1 + \frac{\partial h}{\partial x}$$

I can't delete $h$ on right side. Where did I get wrong?

ohisamadaigaku
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    Are you sure there isn't a typo somewhere with your givens? With the work that I have, I get that the statement can only be true if $h = \frac{h_x}{h_y}$. It's also possible that there's someway that last statement is implied already, I just can't see it. – Ninad Munshi Aug 24 '20 at 01:46
  • @NinadMunshi Yes I've checked a few times and it was correct, how did you lead the conclusion by the way? – ohisamadaigaku Aug 24 '20 at 01:55
  • $$f_x = h + xh_x$$ and $$f_y = 1 +xh_y$$ Based on your username is this from an undergrad multivariable calculus class? – Ninad Munshi Aug 24 '20 at 01:55

1 Answers1

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Use the relationship to rewrite the expression solely in terms of $f$

$$f(x,y) = y + xg(f(x,y))$$

Then take partial derivatives to get the system of equations

$$\begin{cases}\frac{\partial f}{\partial x} = g(f) + x g'(f)\cdot\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} = 1 + x g'(f)\cdot\frac{\partial f}{\partial y} \\ \end{cases} \implies \begin{cases}\frac{\partial f}{\partial x}\left(1-xg'(f)\right) = g(f) = h \\ \frac{\partial f}{\partial y}\left(1-xg'(f)\right) = 1 \\ \end{cases}$$

which means that

$$\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = h \implies \frac{\partial f}{\partial x}(x,y) = h(x,y) \frac{\partial f}{\partial y}(x,y)$$

This may sound like a powerful statement, but in reality there is probably only a very small family of functions that would satisfy those two relationships ($g = $ const is one such trivial relationship)

Ninad Munshi
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