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I saw a similar question here, but there is still something I don't understand:

suppose I have $f = f(x,y,h(x,y))$ then by chain rule $\frac {\partial f}{\partial h}= \frac {\partial f}{\partial x}\frac {\partial x}{\partial h}+\frac {\partial f}{\partial y}\frac {\partial y}{\partial h}$

and if I understand correctly, this equlas to $\frac {\partial f}{\partial x}(\frac {\partial h}{\partial x})^{-1}+\frac {\partial f}{\partial y}(\frac {\partial y}{\partial h})^{-1}$.

But what shall I do if $\frac {\partial x}{\partial h}=0$ or $\frac {\partial y}{\partial h}=0$

Benny K
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    None of your formulas make much sense to me... The first one would be fine if you had something like $f(x(h,k),y(h,k))$, but not with $f(x,y,h(x,y))$. – Hans Lundmark Oct 18 '22 at 07:45
  • @HansLundmark then what is $\frac{\partial f}{\partial h}$ in this case? – Benny K Oct 18 '22 at 08:52
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    It's not at all clear to me what one would even mean by $\partial f/\partial h$ in this case! Could you please clarify what you are actually trying to do with this computation? As it stands, I don't think it's possible to answer your question. – Hans Lundmark Oct 18 '22 at 09:47
  • @HansLundmark for example $f(x,y,h)=x^2-2y+3h^2$ where $h(x,y)=x-y^2$ , what is $\frac{\partial f}{\partial h}$? isn't it $\frac {\partial f}{\partial h}= \frac {\partial f}{\partial x}\frac {\partial x}{\partial h}+\frac {\partial f}{\partial y}\frac {\partial y}{\partial h}$ – Benny K Oct 18 '22 at 09:58
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    If $f(x,y,h)=x^2-2y+3h^2$, where $x$, $y$ and $h$ are independent variables, then $\partial f/\partial h = 0+0+6h$ (of course). But if you start talking about $h(x,y)=x-y^2$, then the question doesn't make sense anymore! What would it mean to take the partial derivative with respect to a function? Or do you want to differentiate the expression $g(x,y) = f(x,y,h(x,y))=x^2-2y+3(x-y^2)^2$? But that's a function of $x$ and $y$ only, not of $h$! – Hans Lundmark Oct 18 '22 at 10:49
  • @HansLundmark $h$ is obviously not independent variable. Can't I write $f=f(x,y,x-y^2)$ and derive it with respect to $x-y^2$? – Benny K Oct 18 '22 at 11:24
  • What would that mean? If you can't explain precisely what you want to do, it's meaningless to ask how to do it. – Hans Lundmark Oct 18 '22 at 13:36

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