How to work out $\left.\frac{\partial f(x,y)}{\partial x}\right\rvert_{h(x,y)}$?

Question

If $f=g(x,y)$, and $h=i(x,y)$, how to work out $$\left.\frac{\partial f}{\partial x}\right\rvert_h $$ in terms of partial derivatives like $\left.\frac{\partial ...}{\partial ...}\right\rvert_x$ and $\left.\frac{\partial ...}{\partial ...}\right\rvert_y$?
I would say $f=g(x,y)=j(x,y,h)$, but it ends up with a correlation for $\left.\frac{\partial f}{\partial x}\right\rvert_{h,y}$.

Answer 1

You wish to take the partial derivative of $g(x,y)$ with respect to $x$ when $i(x,y)$ is the independent variable.

You would need to be able to set $f=j(x,h)$ for some function $j$ such that $\large g(x,y)=j(x,i(x,y))$.

$$\left.\dfrac{\partial f}{\partial x}\right\vert_h = j^{[1,0]}(x,i(x,y))$$

Where $j^{[1,0]}$ s the partial deriative of bivalent function $j$ with respect to its first argument.

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Now, finding such a function might be tricky. But it is essentially what we do in a change of variables, when $u=x, v=i(x,y)$.