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Calculus Volume 1

3.8 Implicit Differentiation

Calculus Volume 13.8 Implicit Differentiation

Learning Objectives

  • 3.8.1 Find the derivative of a complicated function by using implicit differentiation.
  • 3.8.2 Use implicit differentiation to determine the equation of a tangent line.

We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specific point. In all these cases we had the explicit equation for the function and differentiated these functions explicitly. Suppose instead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrary curve at a point. In this section, we solve these problems by finding the derivatives of functions that define yy implicitly in terms of x.x.

Implicit Differentiation

In most discussions of math, if the dependent variable yy is a function of the independent variable x,x, we express y in terms of x.x. If this is the case, we say that yy is an explicit function of x.x. For example, when we write the equation y=x2+1,y=x2+1, we are defining y explicitly in terms of x.x. On the other hand, if the relationship between the function yy and the variable xx is expressed by an equation where yy is not expressed entirely in terms of x,x, we say that the equation defines y implicitly in terms of x.x. For example, the equation yx2=1yx2=1 defines the function y=x2+1y=x2+1 implicitly.

Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of yy are functions that satisfy the given equation, but that yy is not actually a function of x.x.

In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define many different functions implicitly. For example, the functions

y=25x2y=25x2 and y={25x2if5<x<025x2if0<x<25,y={25x2if5<x<025x2if0<x<25, which are illustrated in Figure 3.30, are just three of the many functions defined implicitly by the equation x2+y2=25.x2+y2=25.

The circle with radius 5 and center at the origin is graphed fully in one picture. Then, only its segments in quadrants I and II are graphed. Then, only its segments in quadrants III and IV are graphed. Lastly, only its segments in quadrants II and IV are graphed.
Figure 3.30 The equation x2+y2=25x2+y2=25 defines many functions implicitly.

If we want to find the slope of the line tangent to the graph of x2+y2=25x2+y2=25 at the point (3,4),(3,4), we could evaluate the derivative of the function y=25x2y=25x2 at x=3.x=3. On the other hand, if we want the slope of the tangent line at the point (3,−4),(3,−4), we could use the derivative of y=25x2.y=25x2. However, it is not always easy to solve for a function defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly. The process of finding dydxdydx using implicit differentiation is described in the following problem-solving strategy.

Problem-Solving Strategy

Problem-Solving Strategy: Implicit Differentiation

To perform implicit differentiation on an equation that defines a function yy implicitly in terms of a variable x,x, use the following steps:

  1. Take the derivative of both sides of the equation. Keep in mind that y is a function of x. Consequently, whereas ddx(sinx)=cosx,ddx(siny)=cosydydxddx(sinx)=cosx,ddx(siny)=cosydydx because we must use the chain rule to differentiate sinysiny with respect to x.x.
  2. Rewrite the equation so that all terms containing dydxdydx are on the left and all terms that do not contain dydxdydx are on the right.
  3. Factor out dydxdydx on the left.
  4. Solve for dydxdydx by dividing both sides of the equation by an appropriate algebraic expression.

Example 3.68

Using Implicit Differentiation

Assuming that yy is defined implicitly by the equation x2+y2=25,x2+y2=25, find dydx.dydx.

Analysis

Note that the resulting expression for dydxdydx is in terms of both the independent variable xx and the dependent variable y.y. Although in some cases it may be possible to express dydxdydx in terms of xx only, it is generally not possible to do so.

Example 3.69

Using Implicit Differentiation and the Product Rule

Assuming that yy is defined implicitly by the equation x3siny+y=4x+3,x3siny+y=4x+3, find dydx.dydx.

Example 3.70

Using Implicit Differentiation to Find a Second Derivative

Find d2ydx2d2ydx2 if x2+y2=25.x2+y2=25.

Checkpoint 3.48

Find dydxdydx for yy defined implicitly by the equation 4x5+tany=y2+5x.4x5+tany=y2+5x.

Finding Tangent Lines Implicitly

Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations.

Example 3.71

Finding a Tangent Line to a Circle

Find the equation of the line tangent to the curve x2+y2=25x2+y2=25 at the point (3,−4).(3,−4).

Example 3.72

Finding the Equation of the Tangent Line to a Curve

Find the equation of the line tangent to the graph of y3+x33xy=0y3+x33xy=0 at the point (32,32)(32,32) (Figure 3.32). This curve is known as the folium (or leaf) of Descartes.

A folium is shown, which is a line that creates a loop that crosses over itself. In this graph, it crosses over itself at (0, 0). Its tangent line from (3/2, 3/2) is shown.
Figure 3.32 Finding the tangent line to the folium of Descartes at (32,32).(32,32).

Example 3.73

Applying Implicit Differentiation

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation 4x2+25y2=100.4x2+25y2=100. The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive x-axis toward (0,0).(0,0). If the rocket fires a missile when it is located at (3,85),(3,85), where will it intersect the x-axis?

Checkpoint 3.49

Find the equation of the line tangent to the hyperbola x2y2=16x2y2=16 at the point (5,3).(5,3).

Section 3.8 Exercises

For the following exercises, use implicit differentiation to find dydx.dydx.

300.

x 2 y 2 = 4 x 2 y 2 = 4

301.

6 x 2 + 3 y 2 = 12 6 x 2 + 3 y 2 = 12

302.

x 2 y = y 7 x 2 y = y 7

303.

3 x 3 + 9 x y 2 = 5 x 3 3 x 3 + 9 x y 2 = 5 x 3

304.

x y cos ( x y ) = 1 x y cos ( x y ) = 1

305.

y x + 4 = x y + 8 y x + 4 = x y + 8

306.

x y 2 = x 7 x y 2 = x 7

307.

y sin ( x y ) = y 2 + 2 y sin ( x y ) = y 2 + 2

308.

( x y ) 2 + 3 x = y 2 ( x y ) 2 + 3 x = y 2

309.

x 3 y + x y 3 = −8 x 3 y + x y 3 = −8

For the following exercises, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line.

310.

[T] x4yxy3=−2,(−1,−1)x4yxy3=−2,(−1,−1)

311.

[T] x2y2+5xy=14,(2,1)x2y2+5xy=14,(2,1)

312.

[T] tan(xy)=y,(π4,1)tan(xy)=y,(π4,1)

313.

[T] xy2+sin(πy)2x2=10,(2,−3)xy2+sin(πy)2x2=10,(2,−3)

314.

[T] xy+5x7=34y,(1,2)xy+5x7=34y,(1,2)

315.

[T] xy+sin(x)=1,(π2,0)xy+sin(x)=1,(π2,0)

316.

[T] The graph of a folium of Descartes with equation 2x3+2y39xy=02x3+2y39xy=0 is given in the following graph.

A folium is graphed which has equation 2x3 + 2y3 – 9xy = 0. It crosses over itself at (0, 0).
  1. Find the equation of the tangent line at the point (2,1).(2,1). Graph the tangent line along with the folium.
  2. Find the equation of the normal line to the tangent line in a. at the point (2,1).(2,1).
317.

For the equation x2+2xy3y2=0,x2+2xy3y2=0,

  1. Find the equation of the normal to the tangent line at the point (1,1).(1,1).
  2. At what other point does the normal line in a. intersect the graph of the equation?
318.

Find all points on the graph of y327y=x290y327y=x290 at which the tangent line is vertical.

319.

For the equation x2+xy+y2=7,x2+xy+y2=7,

  1. Find the xx-intercept(s).
  2. Find the slope of the tangent line(s) at the x-intercept(s).
  3. What does the value(s) in b. indicate about the tangent line(s)?
320.

Find the equation of the tangent line to the graph of the equation sin−1x+sin−1y=π6sin−1x+sin−1y=π6 at the point (0,12).(0,12).

321.

Find the equation of the tangent line to the graph of the equation tan−1(x+y)=x2+π4tan−1(x+y)=x2+π4 at the point (0,1).(0,1).

322.

Find yy and yy for x2+6xy2y2=3.x2+6xy2y2=3.

323.

[T] The number of cell phones produced when xx dollars is spent on labor and yy dollars is spent on capital invested by a manufacturer can be modeled by the equation 60x3/4y1/4=3240.60x3/4y1/4=3240.

  1. Find dydxdydx and evaluate at the point (81,16).(81,16).
  2. Interpret the result of a.
324.

[T] The number of cars produced when xx dollars is spent on labor and yy dollars is spent on capital invested by a manufacturer can be modeled by the equation 30x1/3y2/3=360.30x1/3y2/3=360.

(Both xx and yy are measured in thousands of dollars.)

  1. Find dydxdydx and evaluate at the point (27,8).(27,8).
  2. Interpret the result of a.
325.

The volume of a right circular cone of radius xx and height yy is given by V=13πx2y.V=13πx2y. Suppose that the volume of the cone is a constant. Find dydxdydx when x=4x=4 and y=16.y=16.

For the following exercises, consider a closed rectangular box with a square base with side xx and height y.y.

326.

Find an equation for the surface area of the rectangular box, S(x,y).S(x,y).

327.

If the surface area of the rectangular box is 78 square feet, find dydxdydx when x=3x=3 feet and y=5y=5 feet.

For the following exercises, use implicit differentiation to determine y.y. Does the answer agree with the formulas we have previously determined?

328.

x = sin y x = sin y

329.

x = cos y x = cos y

330.

x = tan y x = tan y

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