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Calculus Volume 1

3.8 Implicit Differentiation

Calculus Volume 13.8 Implicit Differentiation
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

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: 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.

Solution

Follow the steps in the problem-solving strategy.

ddx(x2+y2)=ddx(25)Step 1. Differentiate both sides of the equation.ddx(x2)+ddx(y2)=0Step 1.1. Use the sum rule on the left.On the rightddx(25)=0.2x+2ydydx=0Step 1.2. Take the derivatives, soddx(x2)=2xandddx(y2)=2ydydx.2ydydx=−2xStep 2. Keep the terms withdydxon the left.Move the remaining terms to the right.dydx=xyStep 4. Divide both sides of the equation by2y.(Step 3 does not apply in this case.)ddx(x2+y2)=ddx(25)Step 1. Differentiate both sides of the equation.ddx(x2)+ddx(y2)=0Step 1.1. Use the sum rule on the left.On the rightddx(25)=0.2x+2ydydx=0Step 1.2. Take the derivatives, soddx(x2)=2xandddx(y2)=2ydydx.2ydydx=−2xStep 2. Keep the terms withdydxon the left.Move the remaining terms to the right.dydx=xyStep 4. Divide both sides of the equation by2y.(Step 3 does not apply in this case.)

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.

Solution

ddx(x3siny+y)=ddx(4x+3)Step 1: Differentiate both sides of the equation.ddx(x3siny)+ddx(y)=4Step 1.1: Apply the sum rule on the left.On the right,ddx(4x+3)=4.(ddx(x3)·siny+ddx(siny)·x3)+dydx=4Step 1.2: Use the product rule to findddx(x3siny).Observe thatddx(y)=dydx.3x2siny+(cosydydx)·x3+dydx=4Step 1.3: We knowddx(x3)=3x2.Use thechain rule to obtainddx(siny)=cosydydx.x3cosydydx+dydx=43x2sinyStep 2: Keep all terms containingdydxon theleft. Move all other terms to the right.dydx(x3cosy+1)=43x2sinyStep 3: Factor outdydxon the left.dydx=43x2sinyx3cosy+1Step 4: Solve fordydxby dividing both sides ofthe equation byx3cosy+1.ddx(x3siny+y)=ddx(4x+3)Step 1: Differentiate both sides of the equation.ddx(x3siny)+ddx(y)=4Step 1.1: Apply the sum rule on the left.On the right,ddx(4x+3)=4.(ddx(x3)·siny+ddx(siny)·x3)+dydx=4Step 1.2: Use the product rule to findddx(x3siny).Observe thatddx(y)=dydx.3x2siny+(cosydydx)·x3+dydx=4Step 1.3: We knowddx(x3)=3x2.Use thechain rule to obtainddx(siny)=cosydydx.x3cosydydx+dydx=43x2sinyStep 2: Keep all terms containingdydxon theleft. Move all other terms to the right.dydx(x3cosy+1)=43x2sinyStep 3: Factor outdydxon the left.dydx=43x2sinyx3cosy+1Step 4: Solve fordydxby dividing both sides ofthe equation byx3cosy+1.

Example 3.70

Using Implicit Differentiation to Find a Second Derivative

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

Solution

In Example 3.68, we showed that dydx=xy.dydx=xy. We can take the derivative of both sides of this equation to find d2ydx2.d2ydx2.

d2ydx2=ddy(xy)Differentiate both sides ofdydx=xy.=(1·yxdydx)y2Use the quotient rule to findddy(xy).=y+xdydxy2Simplify.=y+x(xy)y2Substitutedydx=xy.=y2x2y3Simplify.d2ydx2=ddy(xy)Differentiate both sides ofdydx=xy.=(1·yxdydx)y2Use the quotient rule to findddy(xy).=y+xdydxy2Simplify.=y+x(xy)y2Substitutedydx=xy.=y2x2y3Simplify.

At this point we have found an expression for d2ydx2.d2ydx2. If we choose, we can simplify the expression further by recalling that x2+y2=25x2+y2=25 and making this substitution in the numerator to obtain d2ydx2=25y3.d2ydx2=25y3.

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).

Solution

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In Example 3.68, we found dydx=xy.dydx=xy.

The slope of the tangent line is found by substituting (3,−4)(3,−4) into this expression. Consequently, the slope of the tangent line is dydx|(3,−4)=3−4=34.dydx|(3,−4)=3−4=34.

Using the point (3,−4)(3,−4) and the slope 3434 in the point-slope equation of the line, we obtain the equation y=34x254y=34x254 (Figure 3.31).

The circle with radius 5 and center at the origin is graphed. A tangent line is drawn through the point (3, −4).
Figure 3.31 The line y=34x254y=34x254 is tangent to x2+y2=25x2+y2=25 at the point (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).

Solution

Begin by finding dydx.dydx.

ddx(y3+x33xy)=ddx(0)3y2dydx+3x2(3y+dydx3x)=0dydx=3y3x23y23x.ddx(y3+x33xy)=ddx(0)3y2dydx+3x2(3y+dydx3x)=0dydx=3y3x23y23x.

Next, substitute (32,32)(32,32) into dydx=3y3x23y23xdydx=3y3x23y23x to find the slope of the tangent line:

dydx|(32,32)=−1.dydx|(32,32)=−1.

Finally, substitute into the point-slope equation of the line to obtain

y=x+3.y=x+3.

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?

Solution

To solve this problem, we must determine where the line tangent to the graph of

4x2+25y2=1004x2+25y2=100 at (3,85)(3,85) intersects the x-axis. Begin by finding dydxdydx implicitly.

Differentiating, we have

8x+50ydydx=0.8x+50ydydx=0.

Solving for dydx,dydx, we have

dydx=4x25y.dydx=4x25y.

The slope of the tangent line is dydx|(3,85)=310.dydx|(3,85)=310. The equation of the tangent line is y=310x+52.y=310x+52. To determine where the line intersects the x-axis, solve 0=310x+52.0=310x+52. The solution is x=253.x=253. The missile intersects the x-axis at the point (253,0).(253,0).

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.

x2y2=4x2y2=4

301.

6x2+3y2=126x2+3y2=12

302.

x2y=y7x2y=y7

303.

3x3+9xy2=5x33x3+9xy2=5x3

304.

xycos(xy)=1xycos(xy)=1

305.

yx+4=xy+8yx+4=xy+8

306.

xy2=x7xy2=x7

307.

ysin(xy)=y2+2ysin(xy)=y2+2

308.

(xy)2+3x=y2(xy)2+3x=y2

309.

x3y+xy3=−8x3y+xy3=−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 85πcm3.85πcm3. 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=sinyx=siny

329.

x=cosyx=cosy

330.

x=tanyx=tany

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