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Calculus Volume 1

3.7 Derivatives of Inverse Functions

Calculus Volume 13.7 Derivatives of Inverse Functions
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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

  • 3.7.1. Calculate the derivative of an inverse function.
  • 3.7.2. Recognize the derivatives of the standard inverse trigonometric functions.

In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents.

The Derivative of an Inverse Function

We begin by considering a function and its inverse. If f(x)f(x) is both invertible and differentiable, it seems reasonable that the inverse of f(x)f(x) is also differentiable. Figure 3.28 shows the relationship between a function f(x)f(x) and its inverse f−1(x).f−1(x). Look at the point (a,f−1(a))(a,f−1(a)) on the graph of f−1(x)f−1(x) having a tangent line with a slope of (f−1)(a)=pq.(f−1)(a)=pq. This point corresponds to a point (f−1(a),a)(f−1(a),a) on the graph of f(x)f(x) having a tangent line with a slope of f(f−1(a))=qp.f(f−1(a))=qp. Thus, if f−1(x)f−1(x) is differentiable at a,a, then it must be the case that

(f−1)(a)=1f(f−1(a)).(f−1)(a)=1f(f−1(a)).
This graph shows a function f(x) and its inverse f−1(x). These functions are symmetric about the line y = x. The tangent line of the function f(x) at the point (f−1(a), a) and the tangent line of the function f−1(x) at (a, f−1(a)) are also symmetric about the line y = x. Specifically, if the slope of one were p/q, then the slope of the other would be q/p. Lastly, their derivatives are also symmetric about the line y = x.
Figure 3.28 The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.

We may also derive the formula for the derivative of the inverse by first recalling that x=f(f−1(x)).x=f(f−1(x)). Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

1=f(f−1(x))(f−1)(x)).1=f(f−1(x))(f−1)(x)).

Solving for (f−1)(x),(f−1)(x), we obtain

(f−1)(x)=1f(f−1(x)).(f−1)(x)=1f(f−1(x)).
3.19

We summarize this result in the following theorem.

Theorem 3.11

Inverse Function Theorem

Let f(x)f(x) be a function that is both invertible and differentiable. Let y=f−1(x)y=f−1(x) be the inverse of f(x).f(x). For all xx satisfying f(f−1(x))0,f(f−1(x))0,

dydx=ddx(f−1(x))=(f−1)(x)=1f(f−1(x)).dydx=ddx(f−1(x))=(f−1)(x)=1f(f−1(x)).

Alternatively, if y=g(x)y=g(x) is the inverse of f(x),f(x), then

g'(x)=1f(g(x)).g'(x)=1f(g(x)).

Example 3.60

Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of g(x)=x+2x.g(x)=x+2x. Compare the resulting derivative to that obtained by differentiating the function directly.

Solution

The inverse of g(x)=x+2xg(x)=x+2x is f(x)=2x1.f(x)=2x1. Since g(x)=1f(g(x)),g(x)=1f(g(x)), begin by finding f(x).f(x). Thus,

f(x)=−2(x1)2andf(g(x))=−2(g(x)1)2=−2(x+2x1)2=x22.f(x)=−2(x1)2andf(g(x))=−2(g(x)1)2=−2(x+2x1)2=x22.

Finally,

g(x)=1f(g(x))=2x2.g(x)=1f(g(x))=2x2.

We can verify that this is the correct derivative by applying the quotient rule to g(x)g(x) to obtain

g(x)=2x2.g(x)=2x2.
Checkpoint 3.42

Use the inverse function theorem to find the derivative of g(x)=1x+2.g(x)=1x+2. Compare the result obtained by differentiating g(x)g(x) directly.

Example 3.61

Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of g(x)=x3.g(x)=x3.

Solution

The function g(x)=x3g(x)=x3 is the inverse of the function f(x)=x3.f(x)=x3. Since g(x)=1f(g(x)),g(x)=1f(g(x)), begin by finding f(x).f(x). Thus,

f(x)=3x3andf(g(x))=3(x3)2=3x2/3.f(x)=3x3andf(g(x))=3(x3)2=3x2/3.

Finally,

g(x)=13x2/3=13x−2/3.g(x)=13x2/3=13x−2/3.
Checkpoint 3.43

Find the derivative of g(x)=x5g(x)=x5 by applying the inverse function theorem.

From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form 1n,1n, where nn is a positive integer. This extension will ultimately allow us to differentiate xq,xq, where qq is any rational number.

Theorem 3.12

Extending the Power Rule to Rational Exponents

The power rule may be extended to rational exponents. That is, if nn is a positive integer, then

ddx(x1/n)=1nx(1/n)1.ddx(x1/n)=1nx(1/n)1.
3.20

Also, if nn is a positive integer and mm is an arbitrary integer, then

ddx(xm/n)=mnx(m/n)1.ddx(xm/n)=mnx(m/n)1.
3.21

Proof

The function g(x)=x1/ng(x)=x1/n is the inverse of the function f(x)=xn.f(x)=xn. Since g(x)=1f(g(x)),g(x)=1f(g(x)), begin by finding f(x).f(x). Thus,

f(x)=nxn1andf(g(x))=n(x1/n)n1=nx(n1)/n.f(x)=nxn1andf(g(x))=n(x1/n)n1=nx(n1)/n.

Finally,

g(x)=1nx(n1)/n=1nx(1n)/n=1nx(1/n)1.g(x)=1nx(n1)/n=1nx(1n)/n=1nx(1/n)1.

To differentiate xm/nxm/n we must rewrite it as (x1/n)m(x1/n)m and apply the chain rule. Thus,

ddx(xm/n)=ddx((x1/n)m)=m(x1/n)m1·1nx(1/n)1=mnx(m/n)1.ddx(xm/n)=ddx((x1/n)m)=m(x1/n)m1·1nx(1/n)1=mnx(m/n)1.

Example 3.62

Applying the Power Rule to a Rational Power

Find the equation of the line tangent to the graph of y=x2/3y=x2/3 at x=8.x=8.

Solution

First find dydxdydx and evaluate it at x=8.x=8. Since

dydx=23x−1/3anddydx|x=8=13dydx=23x−1/3anddydx|x=8=13

the slope of the tangent line to the graph at x=8x=8 is 13.13.

Substituting x=8x=8 into the original function, we obtain y=4.y=4. Thus, the tangent line passes through the point (8,4).(8,4). Substituting into the point-slope formula for a line, we obtain the tangent line

y=13x+43.y=13x+43.

Checkpoint 3.44

Find the derivative of s(t)=2t+1.s(t)=2t+1.

Derivatives of Inverse Trigonometric Functions

We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.

Example 3.63

Derivative of the Inverse Sine Function

Use the inverse function theorem to find the derivative of g(x)=sin−1x.g(x)=sin−1x.

Solution

Since for xx in the interval [π2,π2],f(x)=sinx[π2,π2],f(x)=sinx is the inverse of g(x)=sin−1x,g(x)=sin−1x, begin by finding f(x).f(x). Since

f(x)=cosxandf(g(x))=cos(sin−1x)=1x2,f(x)=cosxandf(g(x))=cos(sin−1x)=1x2,

we see that

g(x)=ddx(sin−1x)=1f(g(x))=11x2.g(x)=ddx(sin−1x)=1f(g(x))=11x2.

Analysis

To see that cos(sin−1x)=1x2,cos(sin−1x)=1x2, consider the following argument. Set sin−1x=θ.sin−1x=θ. In this case, sinθ=xsinθ=x where π2θπ2.π2θπ2. We begin by considering the case where 0<θ<π2.0<θ<π2. Since θθ is an acute angle, we may construct a right triangle having acute angle θ,θ, a hypotenuse of length 11 and the side opposite angle θθ having length x.x. From the Pythagorean theorem, the side adjacent to angle θθ has length 1x2.1x2. This triangle is shown in Figure 3.29. Using the triangle, we see that cos(sin−1x)=cosθ=1x2.cos(sin−1x)=cosθ=1x2.

A right triangle with angle θ, opposite side x, hypotenuse 1, and adjacent side equal to the square root of the quantity (1 – x2).
Figure 3.29 Using a right triangle having acute angle θ,θ, a hypotenuse of length 1,1, and the side opposite angle θθ having length x,x, we can see that cos(sin−1x)=cosθ=1x2.cos(sin−1x)=cosθ=1x2.

In the case where π2<θ<0,π2<θ<0, we make the observation that 0<θ<π20<θ<π2 and hence

cos(sin−1x)=cosθ=cos(θ)=1x2.cos(sin−1x)=cosθ=cos(θ)=1x2.

Now if θ=π2θ=π2 or θ=π2,x=1θ=π2,x=1 or x=−1,x=−1, and since in either case cosθ=0cosθ=0 and 1x2=0,1x2=0, we have

cos(sin−1x)=cosθ=1x2.cos(sin−1x)=cosθ=1x2.

Consequently, in all cases, cos(sin−1x)=1x2.cos(sin−1x)=1x2.

Example 3.64

Applying the Chain Rule to the Inverse Sine Function

Apply the chain rule to the formula derived in Example 3.61 to find the derivative of h(x)=sin−1(g(x))h(x)=sin−1(g(x)) and use this result to find the derivative of h(x)=sin−1(2x3).h(x)=sin−1(2x3).

Solution

Applying the chain rule to h(x)=sin−1(g(x)),h(x)=sin−1(g(x)), we have

h(x)=11(g(x))2g(x).h(x)=11(g(x))2g(x).

Now let g(x)=2x3,g(x)=2x3, so g(x)=6x2.g(x)=6x2. Substituting into the previous result, we obtain

h(x)=114x6·6x2=6x214x6.h(x)=114x6·6x2=6x214x6.
Checkpoint 3.45

Use the inverse function theorem to find the derivative of g(x)=tan−1x.g(x)=tan−1x.

The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.

Theorem 3.13

Derivatives of Inverse Trigonometric Functions

ddxsin−1x=11(x)2ddxsin−1x=11(x)2
3.22
ddxcos−1x=−11(x)2ddxcos−1x=−11(x)2
3.23
ddxtan−1x=11+(x)2ddxtan−1x=11+(x)2
3.24
ddxcot−1x=−11+(x)2ddxcot−1x=−11+(x)2
3.25
ddxsec−1x=1|x|(x)21ddxsec−1x=1|x|(x)21
3.26
ddxcsc−1x=−1|x|(x)21ddxcsc−1x=−1|x|(x)21
3.27

Example 3.65

Applying Differentiation Formulas to an Inverse Tangent Function

Find the derivative of f(x)=tan−1(x2).f(x)=tan−1(x2).

Solution

Let g(x)=x2,g(x)=x2, so g(x)=2x.g(x)=2x. Substituting into Equation 3.24, we obtain

f(x)=11+(x2)2·(2x).f(x)=11+(x2)2·(2x).

Simplifying, we have

f(x)=2x1+x4.f(x)=2x1+x4.

Example 3.66

Applying Differentiation Formulas to an Inverse Sine Function

Find the derivative of h(x)=x2sin−1x.h(x)=x2sin−1x.

Solution

By applying the product rule, we have

h(x)=2xsin−1x+11x2·x2.h(x)=2xsin−1x+11x2·x2.
Checkpoint 3.46

Find the derivative of h(x)=cos−1(3x1).h(x)=cos−1(3x1).

Example 3.67

Applying the Inverse Tangent Function

The position of a particle at time tt is given by s(t)=tan−1(1t)s(t)=tan−1(1t) for t12.t12. Find the velocity of the particle at time t=1.t=1.

Solution

Begin by differentiating s(t)s(t) in order to find v(t).v(t). Thus,

v(t)=s(t)=11+(1t)2·−1t2.v(t)=s(t)=11+(1t)2·−1t2.

Simplifying, we have

v(t)=1t2+1.v(t)=1t2+1.

Thus, v(1)=12.v(1)=12.

Checkpoint 3.47

Find the equation of the line tangent to the graph of f(x)=sin−1xf(x)=sin−1x at x=0.x=0.

Section 3.7 Exercises

For the following exercises, use the graph of y=f(x)y=f(x) to

  1. sketch the graph of y=f−1(x),y=f−1(x), and
  2. use part a. to estimate (f−1)(1).(f−1)(1).
260.
A straight line passing through (0, −3) and (3, 3).
261.
A curved line starting at (−2, 0) and passing through (−1, 1) and (2, 2).
262.
A curved line starting at (4, 0) and passing through (0, 1) and (−1, 4).
263.
A quarter circle starting at (0, 4) and ending at (4, 0).

For the following exercises, use the functions y=f(x)y=f(x) to find

  1. dfdxdfdx at x=ax=a and
  2. x=f−1(y).x=f−1(y).
  3. Then use part b. to find df−1dydf−1dy at y=f(a).y=f(a).
264.

f(x)=6x1,x=−2f(x)=6x1,x=−2

265.

f(x)=2x33,x=1f(x)=2x33,x=1

266.

f(x)=9x2,0x3,x=2f(x)=9x2,0x3,x=2

267.

f(x)=sinx,x=0f(x)=sinx,x=0

For each of the following functions, find (f−1)(a).(f−1)(a).

268.

f(x)=x2+3x+2,x-32,a=2f(x)=x2+3x+2,x-32,a=2

269.

f(x)=x3+2x+3,a=0f(x)=x3+2x+3,a=0

270.

f(x)=x+x,a=2f(x)=x+x,a=2

271.

f(x)=x2x,x<0,a=1f(x)=x2x,x<0,a=1

272.

f(x)=x+sinx,a=0f(x)=x+sinx,a=0

273.

f(x)=tanx+3x2,a=0f(x)=tanx+3x2,a=0

For each of the given functions y=f(x),y=f(x),

  1. find the slope of the tangent line to its inverse function f−1f−1 at the indicated point P,P, and
  2. find the equation of the tangent line to the graph of f−1f−1 at the indicated point.
274.

f(x)=41+x2,P(2,1)f(x)=41+x2,P(2,1)

275.

f(x)=x4,P(2,8)f(x)=x4,P(2,8)

276.

f(x)=(x3+1)4,P(16,1)f(x)=(x3+1)4,P(16,1)

277.

f(x)=x3x+2,P(−8,2)f(x)=x3x+2,P(−8,2)

278.

f(x)=x5+3x34x8,P(−8,1)f(x)=x5+3x34x8,P(−8,1)

For the following exercises, find dydxdydx for the given function.

279.

y=sin−1(x2)y=sin−1(x2)

280.

y=cos−1(x)y=cos−1(x)

281.

y=sec−1(1x)y=sec−1(1x)

282.

y=csc−1xy=csc−1x

283.

y=(1+tan−1x)3y=(1+tan−1x)3

284.

y=cos−1(2x)·sin−1(2x)y=cos−1(2x)·sin−1(2x)

285.

y=1tan−1(x)y=1tan−1(x)

286.

y=sec−1(x)y=sec−1(x)

287.

y=cot−14x2y=cot−14x2

288.

y=x·csc−1xy=x·csc−1x

For the following exercises, use the given values to find (f−1)(a).(f−1)(a).

289.

f(π)=0,f(π)=−1,a=0f(π)=0,f(π)=−1,a=0

290.

f(6)=2,f(6)=13,a=2f(6)=2,f(6)=13,a=2

291.

f(13)=−8,f(13)=2,a=−8f(13)=−8,f(13)=2,a=−8

292.

f(3)=12,f(3)=23,a=12f(3)=12,f(3)=23,a=12

293.

f(1)=−3,f(1)=10,a=−3f(1)=−3,f(1)=10,a=−3

294.

f(1)=0,f(1)=−2,a=0f(1)=0,f(1)=−2,a=0

295.

[T] The position of a moving hockey puck after tt seconds is s(t)=tan−1ts(t)=tan−1t where ss is in meters.

  1. Find the velocity of the hockey puck at any time t.t.
  2. Find the acceleration of the puck at any time t.t.
  3. Evaluate a. and b. for t=2,4,t=2,4, and 66 seconds.
  4. What conclusion can be drawn from the results in c.?
296.

[T] A building that is 225 feet tall casts a shadow of various lengths xx as the day goes by. An angle of elevation θθ is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation dθdxdθdx when x=272x=272 feet.

A building is shown with height 225 ft. A triangle is made with the building height as the opposite side from the angle θ. The adjacent side has length x.
297.

[T] A pole stands 75 feet tall. An angle θθ is formed when wires of various lengths of xx feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle dθdxdθdx when a wire of length 90 feet is attached.

A flagpole is shown with height 75 ft. A triangle is made with the flagpole height as the opposite side from the angle θ. The hypotenuse has length x.
298.

[T] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by θ=tan−1(x2000),θ=tan−1(x2000), where xx is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.

A rocket is shown with in the air with the distance from its nose to the ground being x. A triangle is made with the rocket height as the opposite side from the angle θ. The adjacent side has length 2000.
299.

[T] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle θθ (in radians) given by θ=cot−1x40cot−1x10,θ=cot−1x40cot−1x10,

where xx is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

A person is shown with a right triangle coming from their eye (the right angle being on the opposite side from the eye), with height 10 and base x. There is a line drawn from the eye to the top of the screen, which makes an angle θ with the triangle’s hypotenuse. The screen has a height of 30.
  1. Find dθdx.dθdx.
  2. Evaluate dθdxdθdx for x=5,10,15,x=5,10,15, and 20.
  3. Interpret the results in b..
  4. Evaluate dθdxdθdx for x=25,30,35,x=25,30,35, and 40
  5. Interpret the results in d. At what distance xx should the person stand to maximize his or her viewing angle?
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