 Implicit function theorem

An equation can define a function without giving an explicit formula for it. The implicit function theorem tells us when the constraint $F(x,y)=0$ determines $y$ uniquely from nearby $x$, and makes the resulting function continuously differentiable. It is useful when solving an equation explicitly is cumbersome or impossible: the partial derivatives already reveal whether local recovery is possible.

Let $F$ be a continuously differentiable map from an open subset of ${\mathbb{R}}^m\times {\mathbb{R}}^n$ to ${\mathbb{R}}^n$. Suppose $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map. There are open neighbourhoods $U$ of $x_0$ and $V$ of $y_0$ and a unique continuously differentiable function $g:U\to V$ such that, for $(x,y)\in U\times V$,Thus the nearby level set is a graph of a function. The derivative of this graph isThe matrix inverse acts on the left; the order matters in several dimensions.

For example, take the unit circle, written as the level set $F(x,y)=x^2+y^2-1=0$. At $(x_0,y_0)=(3/5,4/5)$, $F_y=2y_0=8/5\ne 0$, so the local upper branch is $y=g(x)=\sqrt{1-x^2}$. Implicit differentiation givesThis tangent line slope is available directly from the constraint. The negative square root branch is elsewhere on the same circle; it does not destroy uniqueness near the selected upper point.

At $(1,0)$, however, $F_y=0$ and the circle cannot be represented locally as one function $y=g(x)$: just to the left of $x=1$ there are two nearby values of $y$. The circle is still smooth there. Since $F_x=2\ne 0$, use the other coordinate instead: $x=h(y)=\sqrt{1-y^2}$, with $h^{\prime }(0)=0$. A failed coordinate choice is different from a geometric singularity.

The derivative condition is sufficient, not necessary for every possible implicit graph. For $F(x,y)=y^3$, the entire zero set is the smooth graph $y=0$, even though $F_y$ vanishes there. In this case the defining equation itself has a degenerate derivative; the theorem's test gives no conclusion.

The connection to the inverse function theorem is particularly elegant. Augment the constraint to the map $H(x,y)=(x,F(x,y))$, whose Jacobian matrix isIts determinant is $\det D_{y}F$. When this is nonzero, locally invert $H$ and set its second output to zero. The recovered second input is $g(x)$. In this way an implicit constraint becomes an ordinary inverse function problem. ^[ref]

The Implicit Function Theorem is a fundamental result in calculus and differential topology that provides conditions under which a relation defines a function implicitly.