2.8: Implicit Functions

We now turn to the topic of implicit differentiation. We say that $y$ is an implicit function of $x$ if we are given an equation $$\sigma (x, y) = \tau (x, y) \nonumber$$ which determines $y$ as a function of $x$. An example is $x + xy = 2y$. Implicit differentiation is a way of finding the derivative of $y$ without actually solving for $y$ as a function of $x$. Assume that $dy/dx$ exists. The method has two steps:

Step 1: Differentiate both sides of the equation $\sigma(x, y) = \tau (x, y)$ to get a new equation $$\frac{d(\sigma (x, y))}{dx} = \frac{d(\tau(x, y))}{dx}.$$ The Chain Rule is often used in this step.

Step 2: Solve the new Equation $(\PageIndex{1})$ for $dy/dx$. The answer will usually involve both $x$ and $y$.

In each of the examples below, we assume that $dy/dx$ exists and use implicit differentiation to find the value of $dy/dx$.

Example $\PageIndex{1}$

Given the equation $x + xy = 2y$, find $dy/dx$.

Solution

Step 1: $\dfrac{d (x + xy)}{dx} = \dfrac{d(2y)}{dx}$. We find each side by the Sum and Product Rules, $$\begin{align*} \frac{d(x + xy)}{dx} &= \frac{dx + d(xy)}{dx} = \frac{dx + x \ dy + y \ dx}{dx} = 1 + x \frac{dy}{dx} + y. \\ \frac{d(2y)}{dx} &= 2 \frac{dy}{dx} \end{align*}.$$

Thus our new equation is $$1 + x \frac{dy}{dx} + y = 2 \frac{dy}{dx}. \nonumber$$

Step 2: Solve for $dy/dx$. $$\begin{align*} 2 \frac{dy}{dx} - x \frac{dy}{dx} &= 1 + y \\ \frac{dy}{dx} &= \frac{1 + y}{2 - x}. \end{align*}$$

We can check our answer by solving the original equation for $y$ and using ordinary differentiation: $$\begin{align*} x + xy &= 2y \\ 2y - xy &= x. \\ y &= \frac{x}{2 - x}. \end{align*}$$

By the Quotient Rule, $$\frac{dy}{dx} = \frac{(2 - x) \cdot 1 - x(-1)}{(2 - x)^{2}} = \frac{2}{(2 - x)^{2}}. \nonumber$$

A third way to find $dy/dx$ is to solve the original equation for $x$, find $dx/dy$, and then use the Inverse Function Rule. $$\begin{align*} x + xy &= 2y \\ x &= \frac{2y}{1 + y} \\[4pt] \frac{dx}{dy} &= \frac{(1 + y) \cdot 2 - 2y \cdot 1}{(1 + y)^{2}} = \frac{2}{(1 + y)^{2}} \\[4pt] \frac{dy}{dx} &= \frac{1}{2} (1 + y)^{2} \end{align*}$$

To see that our three answers $$\frac{dy}{dx} = \frac{1 + y}{2 - x}, \quad \frac{dy}{dx} = \frac{2}{(2-x)^{2}}, \quad \frac{dy}{dx} = \frac{1}{2} (1+y)^{2} \nonumber$$ are all the same we substitute $\dfrac{x}{2-x}$ for $y$: $$\begin{align*} \frac{dy}{dx} &= \frac{1+y}{2-x} = \frac{1 + \frac{x}{2-x}}{2-x} = \frac{2}{(2-x)^{2}}. \\[4pt] \frac{dy}{dx} &= \frac{1}{2} (1+y)^{2} = \frac{1}{2} \left(1 + \frac{x}{2-x}\right)^{2} = \frac{2}{(2 - x)^{2}} \end{align*}$$

In Example $\PageIndex{1}$, we found $dy/dx$ by three different methods.

(a) Implicit differentiation. We get $dy/dx$ in terms of both $x$ and $y$.
(b) Solve for $y$ as a function of $x$ and differentiate directly. This gives $dy/dx$ in terms of $x$ only.
(c) Solve for $x$ as a function of $y$, find $dx/dy$ directly, and use the Inverse Function Rule. This method gives $dy/dx$ in terms of $y$ only.

Example $\PageIndex{2}$

Given $y + \sqrt{y} = x^{2}$, find $dy/dx$.

Solution

$$\begin{align*} \frac{d(y + \sqrt{y})}{dx} &= \frac{d \left(x^{2}\right)}{dx} \\ \frac{dy}{dx} + \frac{1}{2} y^{-1/2} \frac{dy}{dx} &= 2x \\ \frac{dy}{dx} &= \frac{2x}{1 + \frac{1}{2} y^{-1/2}} \end{align*}$$

This answer can be used to find the slope at any point on the curve. For example, at the point $(\sqrt{2}, 1)$ the slope is $$\frac{2 \sqrt{2}}{1 + \frac{1}{2} \cdot 1^{-1/2}} = \frac{2 \sqrt{2}}{3/2} = \frac{4 \sqrt{2}}{3} \nonumber$$

while at the point $(-\sqrt{2}, 1)$ the slope is $$\frac{2 (-\sqrt{2})}{1 + \frac{1}{2} \cdot 1^{-1/2}} = \frac{-4 \sqrt{2}}{3}. \nonumber$$

To get $dy/dx$ in terms of $x$, we solve the original equation for $y$ using the quadratic formula: $$\begin{align*} y + \sqrt{y} - x^{2} &= 0 \\ \sqrt{y} &= \frac{-1 \pm \sqrt{1 + 4x^{2}}}{2} \end{align*}$$

Since $\sqrt{y} \geq 0$, only one solution may occur: $$\sqrt{y} = \frac{-1 + \sqrt{1 + 4x^{2}}}{2}. \nonumber$$

Then $$y = \left(\frac{-1 + \sqrt{1 + 4x^{2}}}{2}\right)^{2} \nonumber$$

The graph of this function is shown in Figure $\PageIndex{1}$. By substitution we get $$\frac{dy}{dx} = \frac{2x}{1 + \frac{1}{2} y^{-1/2}} = \frac{2x}{1 + \left(-1 + \sqrt{1 + 4x^{2}}\right)^{-1}}. \nonumber$$

Figure $\PageIndex{1}$: Graph of $y + \sqrt{y} = x^{2}$.

Often one side of an implicit function equation is constant and has derivative zero.

Example $\PageIndex{3}$

Given $x^{2} - 2y^{2} = 4$, $y \leq 0$, find $dy/dx$.

Solution

$$\begin{align*} \frac{d \left(x^{2} - 2y^{2}\right)}{dx} &= \frac{d(4)}{dx} \\ \frac{d \left(x^{2} - 2y^{2}\right)}{dx} &= 2x - 4y \frac{dy}{dx} \\ \frac{d(4)}{dx} &= 0 \\ 2x - 4y \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= \frac{-2x}{-4y} = \frac{x}{2y} \end{align*}$$

Solving the original equation for $y$, we get $$\begin{align*} -2y^{2} &= 4 - x^{2}, \quad y \leq 0; \\ y^{2} &= \frac{x^{2} - 4}{2}, \quad y \leq 0; y &= - \sqrt{\frac{x^{2} - 4}{2}}. \end{align*}$$

Thus $dy/dx$ in terms of $x$ is $$\frac{dy}{dx} = \frac{x}{2y} = \frac{x}{-2 \sqrt{\frac{x^{2} - 4}{2}}} = - \frac{x}{2 \left(x^{2} - 4\right)}. \nonumber$$

The graph of this function is shown in Figure $\PageIndex{2}$.

Figure $\PageIndex{2}$: Graph of $x^{2} - 2y^{2} = 4, \quad y \leq 0$.

Implicit differentiation can even be applied to an equation that does not by itself determine $y$ as a function $x$. Sometimes extra inequalities must be assumed in order to make $y$ a function of $x$.

Example $\PageIndex{4}$

Given $$x^{2} + y^{2} = 1,$$ find $dy/dx$. This equation does not determine $y$ as a function of $x$; its graph is the unit circle. Nevertheless we differentiate both sides with respect to $x$ and solve for $dy/dx$, $$2x + 2y \frac{dy}{dx} = 0, \quad \frac{dy}{dx} = -\frac{x}{y}. \nonumber$$

We can conclude that for any system of formulas $S$ which contains the Equation $(\PageIndex{2})$ and also determines $y$ as a function of $x$, it is true that $$\frac{dy}{dx} = -\frac{x}{y}.$$

We can use Equation $(\PageIndex{3})$ to find the slope of the line tangent to the unit circle at any point on the circle. The following examples are illustrated in Figure $\PageIndex{3}$.

Figure $\PageIndex{3}$: The unit circle and several of its tangent lines.

$$\begin{align*} &\frac{dy}{dx} = 0 \text{ at } (0, -1), \quad \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \text{ at } \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \\ &\frac{dy}{dx} = \sqrt{8} \text{ at } \left(\frac{\sqrt{8}}{3}, -\frac{1}{3}\right), \quad \frac{dy}{dx} \text{ is undefined at } (-1, 0). \end{align*}$$

The system of formulas $$x^{2} + y^{2} = 1, \quad y \geq 0 \nonumber$$ gives us $$y = \sqrt{1 - x^{2}}, \quad \frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\sqrt{1 - x^{2}}}. \nonumber$$

Example $\PageIndex{5}$

Find the slope of the line tangent to the curve $$x^{5} y^{3} + xy^{6} = y + 1.$$ at the points $(1, 1)$, $(1, -1)$, and $(0, -1)$.

Solution

The three points are all on the curve, and the first two points have the same $x$ coordinate, so Equation $(\PageIndex{4})$ does not by itself determine $y$ as a function of $x$.

We differentiate with respect to $x$, $$\frac{d \left(x^{5}y^{3} + xy^{6}\right)}{dx} = \frac{d(y+1)}{dx}, \nonumber$$ $$5x^{4} y^{3} + x^{5} \cdot 3y^{2} \ \frac{dy}{dx} + y^{6} + 6xy^{5} \ \frac{dy}{dx} = \frac{dy}{dx}, \nonumber$$

and then solve for $dy/dx$, $$5x^{4} y^{3} + y^{6} + \left(3x^{5} y^{2} + 6xy^{5} - 1\right) \frac{dy}{dx} = 0, \nonumber$$ $$\frac{dy}{dx} = -\frac{5x^{4} y^{3} + y^{6}}{3x^{5} y^{2} + 6xy^{5} - 1}.$$

Substituting, we get $$\begin{align*} \frac{dy}{dx} &= -\frac{6}{8} \text{ at } (1, 1). \\[4pt] \frac{dy}{dx} &= -1 \text{ at } (1, -1). \\[4pt] \frac{dy}{dx} &= +1 \text{ at } (0, -1). \end{align*}$$

Equation $(\PageIndex{5})$ for $dy/dx$ is true of any system $S$ of formulas which contains Equation $(\PageIndex{4})$ and determines $y$ as a function of $x$.

Here is what generally happens in the method of implicit differentiation. Given an equation $$\tau (x, y) = \sigma (x, y)$$ between two terms which may involve the variables $x$ and $y$, we differentiate both sides of the equation and obtain $$\frac{d(\tau (x, y)}{dx} = \frac{d(\sigma (x, y)}{dx}.$$

We then solve Equation $(\PageIndex{7})$ to get $dy/dx$ equal to a term which typically involves both $x$ and $y$. We can conclude that for any system of formulas which contains Equation $(\PageIndex{6})$ and determines $y$ as a function of $x$, Equation $(\PageIndex{7})$ is true. Also, Equation $(\PageIndex{7})$ can be used to find the slope of the tangent line at any point on the curve $\tau(x, y) = \sigma(x, y)$.

Problems for Section 2.8

In Problems 1-26, find $dy/dx$ by implicit differentiation. The answer may involve both $x$ and $y$.

1.	$xy = 1$	2.	$2x^{2} - 3y^{2} = 4, \quad y \leq 0$
3.	$x^{3} + y^{3} = 2$	4.	$x^{3} = y^{5}$
5.	$y = 1/(x + y)$	6.	$y^{2} + 3y - 5 = x$
7.	$x^{-2} + y^{-2} = 1$	8.	$xy^{3} = y + x$
9.	$x^{2} + 3xy + y^{2} = 0$	10.	$x/y + 3y = 2$
11.	$x^{5} = y^{2} - y + 1$	12.	$\sqrt{x} + \sqrt{y} = x + y$
13.	$y = \sqrt{xy + 1}$	14.	$x^{4} + y^{4} = 5$
15.	$xy^{2} - 3x^{2}y + x = 1$	16.	$2xy^{-2} + x^{-2} = y$
17.	$y = \sin (xy)$	18.	$y = \cos (x + y)$
19.	$x = \cos^{2} y$	20.	$x = \sin y + \cos y$
21.	$y = e^{x + 2y}$	22.	$e^{y} = x^{2} + y$
23.	$e^{x} = \ln y$	24.	$\ln y = \sin x$
25.	$y^{2} = \ln (2x + 3y)$	26.	$\ln (\cos y) = 2x + 5$

In Problems 27-33, find the slope of the line tangent to the given curve at the given point or points.

27.	$x^{2} + xy + y^{2} = 7 \text{ at } (1,2) \text{ and } (-1, 3)$.
28.	$x + y^{3} = y \text{ at } (0,0), (0, 1), (0, -1), (-6, 2)$.
29.	$x^{2} - y^{2} = 3 \text{ at } (2,1), (2, -1), (\sqrt{3}, 0)$.
30.	$\tan x = y \text{ at } (\pi/4, 1)$
31.	$2 \sin^{2} x = 3 \cos y \text{ at } (\pi/3, \pi/3)$
32.	$y + e^{y} = 1 + \ln x$ \text{ at } (1, 0)\)
33.	$e^{\sin x} = \ln y \text{ at } (1, 0)$
34.	Given the equation $x^{2} + y^{2} = 1$, find $dy/dx$ and $d^{2}y/dx^{2}$.
35.	Given the equation $2x^{2} - y^{2} = 1$, find $dy/dx$ and $d^{2}y/dx^{2}$.
36.	Differentiating the equation $x^{2} = y^{2}$ implicitly, we get $dy/dx = x/y$. This is undefined at the point $(0, 0)\(. Sketch the graph of the equation to see what happens at the point \((0, 0)$.