# Find the differential of an inverse function of $G(u,v) = (u^4 - u + uv + v^2,\cos u + \sin v)$

I am attempting to solve this problem:

Show that the system of equations \begin{align} x &= u^4 - u + uv + v^2, \\ y &= \cos u + \sin v \end{align} can be solved for $$(u,v)$$ as a smooth function $$F$$ of $$(x,y)$$, in some neighborhood of $$(0,0)$$, in such a way that $$(u,v) = (0,0)$$ when $$(x,y) = (0,1)$$. What is the differential of the resulting function $$F$$ at $$(0,1)$$?

If I define $$G(u,v) = (u^4 - u + uv + v^2,\cos u + \sin v)$$, then $$G^{-1} = F$$. But $$\det dG^{-1}(0,0) = \begin{vmatrix} 0 & 0\\ 0 & 1\\ \end{vmatrix} = 0.$$

This means that I cannot use the Inverse Function Theorem to show $$G^{-1}$$ is smooth near $$(0,0)$$. I suspect that due to the periodic nature of $$y$$, there may be a different inverse function that does allow me to use the Inverse Function Theorem. However, the problem gives me $$F(0,1) = (0,0)$$ so I feel like I should be using $$(u,v)=(0,0)$$ to find the inverse function.

• You made a mistake while calculating the Jacobian of the function $G$: $$d G = \begin{pmatrix} 4u^3-1+v & u + 2v \\ -\sin u & \cos v \end{pmatrix}.$$ Evaluating this at $(u,v) = (0,0)$ we get $$dG(0,0) = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$$ so that $det\,dG^{-1} (0,0) = -1 \neq 0$ and the Inverse Function Theorem applies. – pg_star Mar 5 at 3:06

As pg_star pointed out in the comments, I had calculated the Jacobian of $$G$$ incorrectly. $$dG(0,0) = \begin{pmatrix} -1 & 0\\ 0 & 1\\ \end{pmatrix}$$
Because $$G(0,0) = (0,1)$$, the Inverse Function Theorem says that $$dF(0,1)$$ is the inverse of $$dG(0,0)$$, which is $$\begin{pmatrix} -1 & 0\\ 0 & 1\\ \end{pmatrix} .$$