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 \((0, 0)\).