Let $x=r\cos \theta$ and $y=r\sin \theta$. Find $r_x$.
My answer: $r_x=(r_x)^{-1}=x_r=(\cos \theta)^{-1}$.
Book answer: $$\frac{\partial (r^2)}{\partial x}=\frac{\partial (x^2+y^2)}{\partial x} \implies 2rr_x=2x \implies r_x =\frac{x}{r}=\cos \theta$$
I suspect my answer is incorrect but I don't see why that is the case can anyone offer an explanation?
Where this condition "$r_x$ equals $(r_x)^{-1}$" is coming from?