Is the Inverse Function Theorem Global? I understand the Inverse Function Theorem as follows, suppose we have a function $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$, we can analyse the partial derivatives of any $n$ arguments with respect to each other. So if I have some function $$f(x,y)=e^x+sin(x+ln(y+1))$$
The IVT enables you to derive $\partial x/ \partial y$ and $\partial y/ \partial x$. For example,
$$\frac{\partial y}{ \partial x}=\frac{\partial f/\partial x}{\partial f/\partial y}=\frac{e^x+cos(x+ln(y+1))}{cos(x+ln(y+1))/(y+1)}$$
Is this a global result? Does this tell us $\partial y/ \partial x$ for any $x$?
 A: No, it is local. Here is the rigorous form of the IFT:
Let $V \subseteq \mathbb{R}^n$ be open and $\mathbf{f}: V \to \mathbb{R^n} \text{ be } C^1$  on $V$. If $\mathbf{a} \in V$ and the determinant of the Jacobian Matrix at $a$ is non-zero, then there exists an open set $W \subseteq V$ such that:


*

*$\mathbf{a} \in W$

*$\mathbf{f}$ is injective on W

*$\mathbf{f}^{-1}$ is $C^1$ on $\mathbf{f}(W)$

*for each $\mathbf{y} \in \mathbf{f}(W)$ : $D\mathbf{f}^{-1}(\mathbf{y}) = [D(\mathbf{f})(\mathbf{f}^{-1}(\mathbf{y})]^{-1}$
$\\$
The latter inverse refers to the inverse matrix. So for any given point in the domain that satisfies those properties, it is only guaranteed to hold "near" that point. If those properties are satisfied at every point of the domain then yes, it is global.
EDIT-
Having reviewed your question closer, you seem to be doing implicit-differentiation which I believe (its been awhile so don't take this to the bank) relies on both the chain-rule and the implicit function theorem (which you have also tagged this question with). In that case, the answer is still local, as the implicit function theorem is similarly local.
