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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> $D_yF(x_0,y_0)$ 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 $(x,y)\in U\times V$,
$$
F(x,y)=0\quad\Longleftrightarrow\quad y=g(x),\qquad g(x_0)=y_0.
$$
Thus the nearby <level set> is a <graph of a function>. The <derivative> of this graph is
$$
Dg(x)=-[D_yF(x,g(x))]^{-1}D_xF(x,g(x)).
$$
The <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 $(x_0,y_0)=(3/5,4/5)$, $F_y=2y_0=8/5\ne0$, so the local upper branch is $y=g(x)=\sqrt{1-x^2}$. <Implicit differentiation> gives
$$
g'(x)=-\frac{F_x}{F_y}=-\frac{x}{y},\qquad
\boxed{g'(3/5)=-3/4.}
$$
This <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.

\Image[/implicit-function-theorem-circle-charts.png]
{title=The same circle uses y as a function of x near its upper arc and x as a function of y near its rightmost 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 $F_x=2\ne0$, use the other coordinate instead: $x=h(y)=\sqrt{1-y^2}$, with $h'(0)=0$. 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> is
$$
DH=
\begin{pmatrix}I&0\\D_xF&D_yF\end{pmatrix}.
$$
Its <determinant> is $\det D_yF$. 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. https://www.math.utoronto.ca/courses/mat237y1/20199/notes/Chapter3/S3.3.html[University of Toronto MAT237: the implicit theorem as a corollary of the inverse theorem]{ref}


 Implicit function theorem (source code)