After checking further, this is indeed the case, as when there are two differential numbers this doesn't become a legal operation. The initial statement dydx = rdrdθ is valid, but changing it to dy/dθ=rdr/dx is not. (by this logic this would mean (dx/dr)(1/r) would have to equal dθ/dy, which is...
This expression is the Jacobian which the differentials dydx in cartesians are translated into polar differentials rdrdθ, and I am keeping it in the context of coordinates in a plane where it is valid (as that is where I make the comparison to the proof y'=(r'sinθ+rcosθ)/(r'cosθ-rsinθ), which...
This is a problem that has been bugging me all day. While working with the well-known dydx = rdrdθ, where r is a function of θ I divided both sides of the equation by dxdθ to get dy/dθ = r(dr/dx)
For the left side, I use y = rsinθ and derive with respect to θ to get dy/dθ = sinθdr/dθ + rcosθ...