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]

For two constraints $F(x,y,z)=G(x,y,z)=0$, the implicit function theorem gives differentiable functions $y(x),z(x)$ when the Jacobian determinant $\partial (F,G)/\partial (y,z)$ is nonzero. At its zeros this particular projection can fail even when a geometric curve remains. A nonzero tangent vector with zero $x$ component gives a vertical tangent in that projection. Singular constraint representations require further care: a determinant of zero alone does not prove an infinite derivative or even a regular intersection.

If a holomorphic function $F(z,w)$ satisfies $F(z_0,w_0)=0$ and $F_w(z_0,w_0)\ne 0$, then near $(z_0,w_0)$ its zero set is the graph $w=g(z)$ of a unique holomorphic function. The analogous statement with $F_z\ne 0$ uses $w$ as the local coordinate.

Implicit differentiation differentiates an identity $F(x,y(x))=0$ along a differentiable branch and uses the chain rule to solve for its derivative:The partial derivatives $F_x,F_y$ treat $x,y$ as independent coordinates before evaluating on the branch. The implicit function theorem justifies the branch's local existence and regularity under its hypotheses; differentiating alone does not establish either. For the unit circle at $(3/5,4/5)$, this calculation yields the tangent line $y-4/5=-(3/4)(x-3/5)$.

On a regular level set $F(x,y,z)=0$, holding $z$ fixed gives $F_{x}dx+F_{y}dy=0$. If $F$ is continuously differentiable and both $F_x$ and $F_y$ are nonzero, the implicit function theorem gives $(\partial y/\partial x{)}_z=-F_x/F_y$ and $(\partial x/\partial y{)}_z=-F_y/F_x$. Multiplication proves the identity. The same held-fixed variable is essential, and a zero denominator invalidates the corresponding quotient rather than creating an infinite factor that may be cancelled formally.

On a level set $f(x,y,z)=c$, implicit differentiation gives the three ratios $-f_y/f_x$, $-f_z/f_y$ and $-f_x/f_z$. Their product is $-1$ wherever all the required coordinate representations are differentiable with nonzero denominators. At coordinate singularities the quotient product need not be defined.