I have two functions
$f_1 = \cos(\omega t + \theta)$ and
$f_2 = \sin(\omega t + \theta)$
and want to differentiate $f_1$ for $f_2$ like this:
$f' = \frac{\partial f_1}{\partial f_2}$
Is it correct to do the following:
$f' = \frac{\partial f_1}{\partial f_2}\cdot \frac{\partial x}{\partial x} = \frac{\partial f_1}{\partial x}\cdot \frac{\partial x}{\partial f_2} = \frac{\partial f_1}{\partial x}\cdot (\frac{\partial f_2}{\partial x})^{-1}$
Which would result in $tan(\omega t + \theta)$?
I can't find any proof whether it is correct or not. Also the step where I differentiate the inverse seems to be wrong, since sin(x) is not explicitly invertible, though I could invert it in an area $[-\pi; \pi]$.
Yes, you can do it.
As we know, d(f1)/dx = -wsin(wt + θ) and d(f2)/dx = wcos(wt+ θ)
now you can just divide these two equations, eliminating dx
d(f1)/d(f2) = -tan(wt+θ)
you arent really finding the differential of the inverse here, you are just taking the reciprocal of differentiation of the sin function