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Suppose we want to find the roots of a function $f$ which depends on $x$,$y$ and $r$, i.e. we are solving $f(x,y,r) = 0$. If $r(x,y)$ is a solution to this equation (there doesn't have to be an analytic expression for $r(x,y)$ a priori), can we then say the following about the derivatives of $r$? $$\frac{\partial r(x,y)}{\partial x} = -\frac{\partial f(x,y,r)}{\partial x}\left(\frac{\partial f(x,y,r)}{\partial r}\right)^{-1} \qquad \qquad \qquad \text{similar for $y$}$$

Working out small examples it seems to work, but I am hesitant to really use it.

Geigercounter
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1 Answers1

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I assume your functions take values in $\mathbb R$. Your assumption is $f(x,y,r(x,y)) = 0$ for all $x,y$. If you differentiate with respect to $x$ you get $$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}=0,$$ and similarly for $y$ you find $$\frac{\partial f}{\partial y}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}=0.$$ To get the expression you say, you need $\partial f/\partial r \neq 0$.

Gibbs
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  • That's how I derived it too, through the total derivative. I'm hesitant because it leads to trivial simplifications in a calculation I am doing. But if it works, it works! – Geigercounter Oct 30 '23 at 07:26