Implicit Function Theorem

Implicit Function Theorem: Let $U\subseteq\mathbb{R}^{n+m}$ be open and let $F:U\to\mathbb{R}^m$ be of class $C^1$. Write points as $(x,y)$ with $x\in\mathbb{R}^n$ and $y\in\mathbb{R}^m$. Suppose $(a,b)\in U$ satisfies $F(a,b)=0$ and the $m\times m$ Jacobian matrix with respect to $y$ is invertible at $(a,b)$: $\det\left(\frac{\partial F}{\partial y}(a,b)\right)\neq 0.$ Then there exist neighborhoods $A$ of $a$ and $B$ of $b$ and a unique $C^1$ function $g:A\to B$ such that $F(x,g(x))=0 \quad \text{for all } x\in A.$ Moreover, $g(a)=b$ and its derivative satisfies $Dg(x)= -\left(\frac{\partial F}{\partial y}(x,g(x))\right)^{-1}\left(\frac{\partial F}{\partial x}(x,g(x))\right).$

This theorem formalizes “implicit differentiation” and explains when a level set $F(x,y)=0$ is locally the graph of a smooth function .

Proof sketch: Define $H(x,y)=(x,F(x,y))$ as a map $\mathbb{R}^{n+m}\to\mathbb{R}^{n+m}$. The derivative $DH(a,b)$ is block-triangular with invertible diagonal blocks (identity in $x$ and $\partial F/\partial y$ in $y$), hence invertible. Apply the inverse function theorem to $H$ to solve for $(x,y)$ in terms of $(x,F)$ near $(a,0)$; setting $F=0$ yields $y=g(x)$.