Implicit functions

Some functions are not explicitly defined in the form of $y=f(x)$, but are determined by some relation between the variables $x$ and $y$. Such functions are called implicit functions.

What are implicit functions?

Consider the graph of a univariate function $y=f(x)$. It is a set of points defined as

$$\{(x,y)\in {\mathbb{R}}^2\mid y=f(x)\}.$$

Next, consider the equation $F(x,y)=0$ where $F$ is some bivariate function of $x$ and $y$. Can the set $$\{(x,y)\in {\mathbb{R}}^2\mid F(x,y)=0\}$$ represent the graph of some function? The short answer is "No." For example, consider the unit circle defined by $$F(x,y)=x^2+y^2-1=0.$$ For each $x=a,-1<a<1$, we have the two values $$y=\sqrt{1-a^2}$$ or $$y=-\sqrt{1-a^2}.$$

Therefore, $F(x,y)=0$, or the set

$$\{(x,y)\in {\mathbb{R}}^2\mid F(x,y)=0\},$$

does not define a function of $x$. However, we can define a function based on a subset of the above set. For instance, the subset

$$\{(x,y)\in {\mathbb{R}}^2\mid F(x,y)=0,y>0\}$$

defines the function $\phi (x)=\sqrt{1-x^2}$ on the open interval $(-1,1)$. A part of a curve (such as $F(x,y)=0$) is called a branch. We can take another branch ($y<0$) to define another function $\phi (x)=-\sqrt{1-x^2}.$

Let's examine this idea more carefully. Let $P=(a,b)$ be a point on the curve $F(x,y)=x^2+y^2-1=0$ (so that $a^2+b^2=1$ holds). If $b\ne 0$, then the point $P$ is either on the branch above the $x$-axis or on the branch below the $x$-axis. In this case, we can take the branch that passes through $P$ and determine a function $y=\phi (x)$ on a neighbor $I$ of $x=a$ (e.g., $I=(-1,1)$). This function $y=\phi (x)$ has the following properties.

1. For all $x\in I$, $F(x,\phi (x))=0$.
2. $b=\phi (a)$.

The property (1) indicates that the graph of $y=\phi (x)$ matches a branch of the curve $F(x,y)=0$. The property (1) indicates that the branch above passes through the point $P=(a,b)$.

If the point $P(a,b)$ satisfies $b=0$ (i.e., $P=(\pm 1,0)$ in the case of the unit circle), then we cannot define a function from any branch that passes through $P$. In fact, if we take any $x<1$ (however close $x$ is to 1), there are always two values of $y$ that satisfy $F(x,y)=0$ so we cannot define a function in the neighbor of the point $(1,0)$. Note that $F_y(x,y)=2y$ so that $b=0$ is equivalent to $F_y(a,b)=0$. The reason why we cannot define a function around $P(a,b)$ when $b=0$ is that the tangent line at $P$ becomes vertical (perpendicular to the $x$-axis).

Definition (Implicit function)

Given the equation $F(x,y)=0$, the univariate function $y=\phi (x)$ is said to be an implicit function of $F(x,y)=0$ if

$$F(x,\phi (x))=0$$

for all $x\in \text{dom}\phi$.

Example. The functions $y=\sqrt{1-x^2}$ and $y=-\sqrt{1-x^2}$ on the open interval $(-1,1)$ are implicit functions of $x^2+y^2-1=0$. □

Implicit function theorem

The above observation on the unit circle can be generalized to the following theorem.

Theorem (Implicit function theorem)

Let $F(x,y)$ be a function of class $C^1$ on an open region $U\subset {\mathbb{R}}^2$. Suppose the point $P=(a,b)$ satisfies the following conditions:

• $F(a,b)=0$ (i.e., $P$ is a point on the curve $F(x,y)=0$),
• $F_y(a,b)\ne 0$.

Then, there exist an open interval $I$ on the $x$-axis such that $a\in I$ and a univariate function $y=\phi (x)$ on $I$ such that

1. $F(x,\phi (x))=0$ for all $x\in I$, that is, $\phi (x)$  is an implicit function of $F(x,y)=0$, and
2. $b=\phi (a)$.

Furthermore, the function $\phi (x)$ is differentiable on $I$ and

$$\begin{array}{}\text{(Eq:IF)}& {\phi }^{\prime }(x)=-\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}.\end{array}$$

Proof. See The Implicit Function Theorem: A proof. ■

Remark. Provided the existence of the implicit function $\phi (x)$, (Eq:IF) can be derived as follows. Differentiating the both sides of $F(x,\phi (x))=0$ with respect to $x$, we have

$$F_x(x,\phi (x))+F_y(x,\phi (x)){\phi }^{\prime }(x)=0.$$

Solving for ${\phi }^{\prime }(x)$, we obtain (Eq:IF). □

Let us examine the implication of the theorem (Figure 1 below).

Figure 1. The graph of an implicit function $y=\phi (x)$.

Suppose the curve $F(x,y)=0$ is given as in Figure 1. If the point $P=(a,b)$ on this curve satisfies $F_y(a,b)\ne 0$ (i.e., the slope of the tangent is not vertical), then we can take some open interval $I$ that includes $x=a$ and a branch that includes $P$. Then this branch defines a function $\phi (x)$ on $I$. On the other hand, if $F_y(a,b)=0$ as in the case for the point $Q$ in Figure 1, we cannot define any implicit functions around that point.

In general, a curve $F(x,y)=0$ may not be a function of $x$, but we may be able to "carve out" a branch that is a function of $x$. This is what the Implicit Function Theorem states.

Remark.

1. Roughly speaking, the implicit function $y=\phi (x)$ is a solution to the equation $F(x,y)=0$ when solved for $y$. However, an explicit "solution" may be impossible to obtain in many cases. Nevertheless, what's important about implicit functions is their existence. In many cases, we don't need to explicitly solve for the function. (After all, they are implicit functions!)
2. When $F_y(a,b)=0$, we may apply the theorem by swapping the roles of $x$ and $y$ so that there may exist a function of $y$ such as $x=\psi (y)$ if $F_x(a,b)\ne 0$. However, if $F_x(a,b)=F_y(a,b)=0$, then we cannot apply this approach either. If the point $P=(a,b)$ satisfies $F_x(a,b)=F_y(a,b)=0$ on the curve $F(x,y)=0$, then $P$ is called a singular point. The points that are not singular are called regular points.

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Application: Tangent of a curve

Using the Implicit Function Theorem, we may obtain the tangent of a general curve  $F(x,y)=0$ which may not be the graph of a function.

Suppose $(a,b)$ is a regular point on the curve $F(x,y)=0$ (i.e., $F(a,b)=0$, and not $F_x(a,b)=F_y(a,b)=0$). Then, if, for example, $F_y(a,b)\ne 0$, then the shape of the curve in a neighbor of $(a,b)$ matches the graph of the implicit function $y=\phi (x)$. Therefore, the equation of the tangent of $F(x,y)=0$ at $(a,b)$ is given by

$$y-b={\phi }^{\prime }(a)(x-a)$$

or

$$F_x(a,b)(x-a)+F_y(a,b)(y-b)=0.$$

In the case when $F_x(a,b)\ne 0$ (and possibly $F_y(a,b)=0$), we obtain the same equation.

Example. Let us find the tangent line of the curve $F(x,y)=y^2-x^3+x=0$ at $(2,\sqrt{6})$. We have

$$\begin{array}{rcl}{F}_x(x,y)& =& -3x^2+1,\\ F_y(x,y)& =& 2y\end{array}$$

so that

$$\begin{array}{rcl}{F}_x(2,\sqrt{6})& =& -11,\\ F_y(2,\sqrt{6})& =& 2\sqrt{6}.\end{array}$$

Thus, the equation of the tangent line is given by

$$-11(x-2)+2\sqrt{6}(y-\sqrt{6})=0$$

or

$$11x-2\sqrt{6}y-10=0.$$

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