4 ms·
Seems like both of you essentially apply the implicit function theorem https://en.wikipedia.org/wiki/Implicit_function_theorem https://en.wikipedia.org/wiki/Im
by zwaps 5y ago
Seems like both of you essentially apply the implicit function theorem
https://en.wikipedia.org/wiki/Implicit_function_theorem https://en.wikipedia.org/wiki/Implicit_function_theorem
I always found that theorem very useful, not only for computing implicit derivatives, but also for understanding if such things actually exist.
Another way to see it is to use the total derivative and then (somewhat heuristically) solve it for dy/dx. Mathematicians may yell at you, tho
- Tainnor 5y agoSort of. Although, for the part that I showed, you don't need the extra assumptions of the implicit function theorem. You only need local invertibility once you actually want to solve the equation. But analysis has been a while for me. edit: to be more precise, nothing in my answer requires knowing the implicit function theorem. I'm just basically executing the same steps as for the proof of the implicit function theorem, but only for the concrete case of f: R^2 -> R (the implicit function theorem works for any f: R^n -> R^m). If you solve the equation, you need local invertibility (in this concrete example: the partial derivative of f wrt. y needs to be invertible in a neighbourhood of (x, y(x)). > Another way to see it is to use the total derivative and then (somewhat heuristically) solve it for dy/dx. Mathematicians may yell at you, tho If you do it properly (i.e. concretely state your assumptions), I don't think mathematicians will yell at you