In multivariable calculus, a branch of mathematics, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.

Example

---

Consider the unit circle. If we define the function $f$ as $f(x,y)\; =\; x^2\; +\; y^2\; -\; 1$, then the relation $f(x,y)\; =\; 0$ cuts out the unit circle. Explicitly, the unit circle is the set $\backslash \{\; (x,y)\; |\; f(x,y)\; =\; 0\; \backslash \}$. There is no way to represent the unit circle as the graph of a function $y\; =\; g(x)$ because for each non-zero choice of $x$, there are two choices of $y$, namely $\backslash sqrt\{1-x^2\}$ and $-\backslash sqrt\{1-x^2\}$.

However, it is possible to represent part of the circle as a function. If we let $g\_1(x)\; =\; \backslash sqrt\{1-x^2\}$ for , then the graph of $y\; =\; g\_1(x)$ gives the upper half of the circle. Similarly, if $g\_2(x)\; =\; -\backslash sqrt\{1-x^2\}$, then the graph of $y\; =\; g\_2(x)$ gives the lower half of the circle.

It is not possible to find a function which will cut out a neighboorhood of $(1,0)$ or $(-1,0)$. Any neighboorhood of $(1,0)$ or $(-1,0)$ contains both the upper and lower halves of the circle. Because functions must be single-valued, there is no way of writing both the upper and lower halves using one function $y\; =\; g(x)$. Consequently there is no function whose graph looks like a neighboorhood of $(1,0)$ or $(-1,0)$. In this case the conclusion of the implicit function theorem fails.

The purpose of the implicit function theorem is to tell us the existence of functions like $g\_1(x)$ and $g\_2(x)$ in situations where we cannot write down explicit formulas. It guarantees that $g\_1(x)$ and $g\_2(x)$ are differentiable, and it even works in situations where we do not have a formula for $f(x,y)$.

Statement of the theorem

---

Let $f\; :\; \backslash bold\; R^\{n+m\}\; \backslash rightarrow\; \backslash bold\; R^n$ be a continuously differentiable function. We think of $\backslash bold\; R^\{n+m\}$ as the cartesian product $\backslash bold\; R^n\; \backslash times\; \backslash bold\; R^m$, and we write a point of this product as $(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)$. $f$ is the given relation. Our goal is to construct a function $g\; :\; \backslash bold\; R^n\; \backslash rightarrow\; \backslash bold\; R^m$ whose graph is the set of all $(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)$ such that $f(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)\; =\; 0$.

As noted above, this may not always be possible, so we will fix a point $(a\_1,\; \backslash ldots\; a\_n,\; b\_1,\; \backslash ldots,\; b\_m)$ which satisfies $f(a\_1,\; \backslash ldots\; a\_n,\; b\_1,\; \backslash ldots,\; b\_m)\; =\; 0$, and we will ask for a $g$ that works near the point $(a\_1,\; \backslash ldots\; a\_n,\; b\_1,\; \backslash ldots,\; b\_m)$. In other words, we want an open set $U$ of $\backslash bold\; R^n$, an open set $V$ of $\backslash bold\; R^m$, and a function $g\; :\; U\; \backslash rightarrow\; V$ such that the graph of $g$ equals the relation $f\; =\; 0$ on $U\; \backslash times\; V$. In symbols,

$\backslash \{\; (x\_1,\; \backslash ldots,\; x\_n,\; g(x\_1,\; \backslash ldots\; x\_n))\; \backslash \}\; =\; \backslash \{\; (x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)\; |\; f(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)\; =\; 0\; \backslash \}\; \backslash cap\; (U\; \backslash times\; V)$

To state the implicit function theorem, we need the Jacobian, also called the differential or total derivative, of $f$. This is the matrix of partial derivatives of $f$. Abbreviating $(a\_1,\; \backslash ldots,\; a\_n,\; b\_1,\; \backslash ldots\; b\_m)$ to $(a,\; b)$, the Jacobian matrix is

$\backslash begin\{matrix\}$

(Df)(a,b) & = & \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(a,b) & \cdots & \frac{\partial f_1}{\partial x_n}(a,b) & \frac{\partial f_1}{\partial y_1}(a,b) & \cdots & \frac{\partial f_1}{\partial y_m}(a,b)\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ \frac{\partial f_n}{\partial x_1}(a,b) & \cdots & \frac{\partial f_n}{\partial x_n}(a,b) & \frac{\partial f_n}{\partial y_1}(a,b) & \cdots & \frac{\partial f_n}{\partial y_m}(a,b)\\ \end{bmatrix}\\ & = & \begin{bmatrix} X & | & Y \end{bmatrix}\\ \end{matrix}

where $X$ is the matrix of partial derivatives in the $x$'s and $Y$ is the matrix of partial derivatives in the $y$'s. The implicit function theorem says that if $X$ is an invertible matrix, then there are $U$, $V$, and $g$ as desired. Writing all the hypotheses together gives the following statement.

Let $f\; :\; \backslash bold\; R^\{n+m\}\; \backslash rightarrow\; \backslash bold\; R^n$ be a continuously differentiable function, and let $\backslash bold\; R^\{n+m\}$ have coordinates $(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)$. Fix a point $(a\_1,\; \backslash ldots\; a\_n,\; b\_1,\; \backslash ldots,\; b\_m)\; =\; (a,b)$. If the matrix $\backslash begin\{bmatrix\}\; (\backslash partial\; f\_i\; /\; \backslash partial\; x\_j)(a,b)\; \backslash end\{bmatrix\}$ is invertible, then there exists an open set $U$ containing $(a\_1,\; \backslash ldots,\; a\_n)$, an open set $V$ containing $(b\_1,\; \backslash ldots,\; b\_m)$, and a differentiable function $g\; :\; U\; \backslash rightarrow\; V$ such that $\backslash \{\; (x\_1,\; \backslash ldots,\; x\_n,\; g(x\_1,\; \backslash ldots\; x\_n))\; \backslash \}\; =\; \backslash \{\; (x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)\; |\; f(x\_1,\; \backslash ldots\; x\_n,\; y\_1,\; \backslash ldots\; y\_m)\; =\; 0\; \backslash \}\; \backslash cap\; (U\; \backslash times\; V)$.

Remark

---

The condition of the implicit function theorem is sufficient, but not necessary. For example, $f(x)\; =\; x^3$ has the inverse $g(x)\; =\; x^\{1/3\}$, even though $f\text{'}(0)\; =\; 0$.

References

---

Mathematical theorems | Multivariate calculus | Differential topology

Satz von der impliziten Funktion | Teorema delle funzioni implicite | משפט הפונקציות הסתומות | Теорема о неявной функции