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Calculus Volume 1

3.9 Derivatives of Exponential and Logarithmic Functions

Calculus Volume 13.9 Derivatives of Exponential and Logarithmic 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

  • 3.9.1. Find the derivative of exponential functions.
  • 3.9.2. Find the derivative of logarithmic functions.
  • 3.9.3. Use logarithmic differentiation to determine the derivative of a function.

So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.

Derivative of the Exponential Function

Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.

First of all, we begin with the assumption that the function B(x)=bx,b>0,B(x)=bx,b>0, is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of bn,bn, where nn is a positive integer—as the product of bb multiplied by itself nn times. Later, we defined b0=1,bn=1bn,b0=1,bn=1bn, for a positive integer n,n, and bs/t=(bt)sbs/t=(bt)s for positive integers ss and t.t. These definitions leave open the question of the value of brbr where rr is an arbitrary real number. By assuming the continuity of B(x)=bx,b>0,B(x)=bx,b>0, we may interpret brbr as limxrbxlimxrbx where the values of xx as we take the limit are rational. For example, we may view 4π4π as the number satisfying

43<4π<44,43.1<4π<43.2,43.14<4π<43.15,43.141<4π<43.142,43.1415<4π<43.1416,.43<4π<44,43.1<4π<43.2,43.14<4π<43.15,43.141<4π<43.142,43.1415<4π<43.1416,.

As we see in the following table, 4π77.88.4π77.88.

xx 4x4x xx 4x4x
4343 64 43.14159343.141593 77.8802710486
43.143.1 73.5166947198 43.141643.1416 77.8810268071
43.1443.14 77.7084726013 43.14243.142 77.9242251944
43.14143.141 77.8162741237 43.1543.15 78.7932424541
43.141543.1415 77.8702309526 43.243.2 84.4485062895
43.1415943.14159 77.8799471543 4444 256
Table 3.6 Approximating a Value of 4π4π

We also assume that for B(x)=bx,b>0,B(x)=bx,b>0, the value B(0)B(0) of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function B(x)B(x) is differentiable everywhere.

We make one final assumption: that there is a unique value of b>0b>0 for which B(0)=1.B(0)=1. We define ee to be this unique value, as we did in Introduction to Functions and Graphs. Figure 3.33 provides graphs of the functions y=2x,y=3x,y=2.7x,y=2x,y=3x,y=2.7x, and y=2.8x.y=2.8x. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function E(x)=exE(x)=ex is called the natural exponential function. Its inverse, L(x)=logex=lnxL(x)=logex=lnx is called the natural logarithmic function.

The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).
Figure 3.33 The graph of E(x)=exE(x)=ex is between y=2xy=2x and y=3x.y=3x.

For a better estimate of e,e, we may construct a table of estimates of B(0)B(0) for functions of the form B(x)=bx.B(x)=bx. Before doing this, recall that

B(0)=limx0bxb0x0=limx0bx1xbx1xB(0)=limx0bxb0x0=limx0bx1xbx1x

for values of xx very close to zero. For our estimates, we choose x=0.00001x=0.00001 and x=−0.00001x=−0.00001 to obtain the estimate

b−0.000011−0.00001<B(0)<b0.0000110.00001.b−0.000011−0.00001<B(0)<b0.0000110.00001.

See the following table.

bb b−0.000011−0.00001<B(0)<b0.0000110.00001b−0.000011−0.00001<B(0)<b0.0000110.00001 bb b−0.000011−0.00001<B(0)<b0.0000110.00001b−0.000011−0.00001<B(0)<b0.0000110.00001
22 0.693145<B(0)<0.693150.693145<B(0)<0.69315 2.71832.7183 1.000002<B(0)<1.0000121.000002<B(0)<1.000012
2.72.7 0.993247<B(0)<0.9932570.993247<B(0)<0.993257 2.7192.719 1.000259<B(0)<1.0002691.000259<B(0)<1.000269
2.712.71 0.996944<B(0)<0.9969540.996944<B(0)<0.996954 2.722.72 1.000627<B(0)<1.0006371.000627<B(0)<1.000637
2.7182.718 0.999891<B(0)<0.9999010.999891<B(0)<0.999901 2.82.8 1.029614<B(0)<1.0296251.029614<B(0)<1.029625
2.71822.7182 0.999965<B(0)<0.9999750.999965<B(0)<0.999975 33 1.098606<B(0)<1.0986181.098606<B(0)<1.098618
Table 3.7 Estimating a Value of ee

The evidence from the table suggests that 2.7182<e<2.7183.2.7182<e<2.7183.

The graph of E(x)=exE(x)=ex together with the line y=x+1y=x+1 are shown in Figure 3.34. This line is tangent to the graph of E(x)=exE(x)=ex at x=0.x=0.

Graph of the function ex along with its tangent at (0, 1), x + 1.
Figure 3.34 The tangent line to E(x)=exE(x)=ex at x=0x=0 has slope 1.

Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of B(x)=bx,b>0.B(x)=bx,b>0. Recall that we have assumed that B(0)B(0) exists. By applying the limit definition to the derivative we conclude that

B(0)=limh0b0+hb0h=limh0bh1h.B(0)=limh0b0+hb0h=limh0bh1h.
3.28

Turning to B(x),B(x), we obtain the following.

B(x)=limh0bx+hbxhApply the limit definition of the derivative.=limh0bxbhbxhNote thatbx+h=bxbh.=limh0bx(bh1)hFactor outbx.=bxlimh0bh1hApply a property of limits.=bxB(0)UseB(0)=limh0b0+hb0h=limh0bh1h.B(x)=limh0bx+hbxhApply the limit definition of the derivative.=limh0bxbhbxhNote thatbx+h=bxbh.=limh0bx(bh1)hFactor outbx.=bxlimh0bh1hApply a property of limits.=bxB(0)UseB(0)=limh0b0+hb0h=limh0bh1h.

We see that on the basis of the assumption that B(x)=bxB(x)=bx is differentiable at 0,B(x)0,B(x) is not only differentiable everywhere, but its derivative is

B(x)=bxB(0).B(x)=bxB(0).
3.29

For E(x)=ex,E(0)=1.E(x)=ex,E(0)=1. Thus, we have E(x)=ex.E(x)=ex. (The value of B(0)B(0) for an arbitrary function of the form B(x)=bx,b>0,B(x)=bx,b>0, will be derived later.)

Theorem 3.14

Derivative of the Natural Exponential Function

Let E(x)=exE(x)=ex be the natural exponential function. Then

E(x)=ex.E(x)=ex.

In general,

ddx(eg(x))=eg(x)g(x).ddx(eg(x))=eg(x)g(x).

Example 3.74

Derivative of an Exponential Function

Find the derivative of f(x)=etan(2x).f(x)=etan(2x).

Solution

Using the derivative formula and the chain rule,

f(x)=etan(2x)ddx(tan(2x))=etan(2x)sec2(2x)·2.f(x)=etan(2x)ddx(tan(2x))=etan(2x)sec2(2x)·2.

Example 3.75

Combining Differentiation Rules

Find the derivative of y=ex2x.y=ex2x.

Solution

Use the derivative of the natural exponential function, the quotient rule, and the chain rule.

y=(ex2·2)x·x1·ex2x2Apply the quotient rule. =ex2(2x21)x2Simplify.y=(ex2·2)x·x1·ex2x2Apply the quotient rule. =ex2(2x21)x2Simplify.
Checkpoint 3.50

Find the derivative of h(x)=xe2x.h(x)=xe2x.

Example 3.76

Applying the Natural Exponential Function

A colony of mosquitoes has an initial population of 1000. After tt days, the population is given by A(t)=1000e0.3t.A(t)=1000e0.3t. Show that the ratio of the rate of change of the population, A(t),A(t), to the population, A(t)A(t) is constant.

Solution

First find A(t).A(t). By using the chain rule, we have A(t)=300e0.3t.A(t)=300e0.3t. Thus, the ratio of the rate of change of the population to the population is given by

A(t)=300e0.3t1000e0.3t=0.3.A(t)=300e0.3t1000e0.3t=0.3.

The ratio of the rate of change of the population to the population is the constant 0.3.

Checkpoint 3.51

If A(t)=1000e0.3tA(t)=1000e0.3t describes the mosquito population after tt days, as in the preceding example, what is the rate of change of A(t)A(t) after 4 days?

Derivative of the Logarithmic Function

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

Theorem 3.15

The Derivative of the Natural Logarithmic Function

If x>0x>0 and y=lnx,y=lnx, then

dydx=1x.dydx=1x.
3.30

More generally, let g(x)g(x) be a differentiable function. For all values of xx for which g(x)>0,g(x)>0, the derivative of h(x)=ln(g(x))h(x)=ln(g(x)) is given by

h(x)=1g(x)g(x).h(x)=1g(x)g(x).
3.31

Proof

If x>0x>0 and y=lnx,y=lnx, then ey=x.ey=x. Differentiating both sides of this equation results in the equation

eydydx=1.eydydx=1.

Solving for dydxdydx yields

dydx=1ey.dydx=1ey.

Finally, we substitute x=eyx=ey to obtain

dydx=1x.dydx=1x.

We may also derive this result by applying the inverse function theorem, as follows. Since y=g(x)=lnxy=g(x)=lnx is the inverse of f(x)=ex,f(x)=ex, by applying the inverse function theorem we have

dydx=1f(g(x))=1elnx=1x.dydx=1f(g(x))=1elnx=1x.

Using this result and applying the chain rule to h(x)=ln(g(x))h(x)=ln(g(x)) yields

h(x)=1g(x)g(x).h(x)=1g(x)g(x).

The graph of y=lnxy=lnx and its derivative dydx=1xdydx=1x are shown in Figure 3.35.

Graph of the function ln x along with its derivative 1/x. The function ln x is increasing on (0, + ∞). Its derivative is decreasing but greater than 0 on (0, + ∞).
Figure 3.35 The functiony=lnxThe functiony=lnx is increasing on (0,+).(0,+). Its derivative y=1xy=1x is greater than zero on (0,+).(0,+).

Example 3.77

Taking a Derivative of a Natural Logarithm

Find the derivative of f(x)=ln(x3+3x4).f(x)=ln(x3+3x4).

Solution

Use Equation 3.31 directly.

f(x)=1x3+3x4·(3x2+3)Useg(x)=x3+3x4inh(x)=1g(x)g(x).=3x2+3x3+3x4Rewrite.f(x)=1x3+3x4·(3x2+3)Useg(x)=x3+3x4inh(x)=1g(x)g(x).=3x2+3x3+3x4Rewrite.

Example 3.78

Using Properties of Logarithms in a Derivative

Find the derivative of f(x)=ln(x2sinx2x+1).f(x)=ln(x2sinx2x+1).

Solution

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

f(x)=ln(x2sinx2x+1)=2lnx+ln(sinx)ln(2x+1)Apply properties of logarithms.f(x)=2x+cotx22x+1Apply sum rule andh(x)=1g(x)g(x).f(x)=ln(x2sinx2x+1)=2lnx+ln(sinx)ln(2x+1)Apply properties of logarithms.f(x)=2x+cotx22x+1Apply sum rule andh(x)=1g(x)g(x).

Checkpoint 3.52

Differentiate: f(x)=ln(3x+2)5.f(x)=ln(3x+2)5.

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y=logbxy=logbx and y=bxy=bx for b>0,b1.b>0,b1.

Theorem 3.16

Derivatives of General Exponential and Logarithmic Functions

Let b>0,b1,b>0,b1, and let g(x)g(x) be a differentiable function.

  1. If, y=logbx,y=logbx, then
    dydx=1xlnb.dydx=1xlnb.
    3.32

    More generally, if h(x)=logb(g(x)),h(x)=logb(g(x)), then for all values of x for which g(x)>0,g(x)>0,
    h(x)=g(x)g(x)lnb.h(x)=g(x)g(x)lnb.
    3.33
  2. If y=bx,y=bx, then
    dydx=bxlnb.dydx=bxlnb.
    3.34

    More generally, if h(x)=bg(x),h(x)=bg(x), then
    h(x)=bg(x)g(x)lnb.h(x)=bg(x)g(x)lnb.
    3.35

Proof

If y=logbx,y=logbx, then by=x.by=x. It follows that ln(by)=lnx.ln(by)=lnx. Thus ylnb=lnx.ylnb=lnx. Solving for y,y, we have y=lnxlnb.y=lnxlnb. Differentiating and keeping in mind that lnblnb is a constant, we see that

dydx=1xlnb.dydx=1xlnb.

The derivative in Equation 3.33 now follows from the chain rule.

If y=bx,y=bx, then lny=xlnb.lny=xlnb. Using implicit differentiation, again keeping in mind that lnblnb is constant, it follows that 1ydydx=lnb.1ydydx=lnb. Solving for dydxdydx and substituting y=bx,y=bx, we see that

dydx=ylnb=bxlnb.dydx=ylnb=bxlnb.

The more general derivative (Equation 3.35) follows from the chain rule.

Example 3.79

Applying Derivative Formulas

Find the derivative of h(x)=3x3x+2.h(x)=3x3x+2.

Solution

Use the quotient rule and Derivatives of General Exponential and Logarithmic Functions.

h(x)=3xln3(3x+2)3xln3(3x)(3x+2)2Apply the quotient rule.=2·3xln3(3x+2)2Simplify.h(x)=3xln3(3x+2)3xln3(3x)(3x+2)2Apply the quotient rule.=2·3xln3(3x+2)2Simplify.

Example 3.80

Finding the Slope of a Tangent Line

Find the slope of the line tangent to the graph of y=log2(3x+1)y=log2(3x+1) at x=1.x=1.

Solution

To find the slope, we must evaluate dydxdydx at x=1.x=1. Using Equation 3.33, we see that

dydx=3ln2(3x+1).dydx=3ln2(3x+1).

By evaluating the derivative at x=1,x=1, we see that the tangent line has slope

dydx|x=1=34ln2=3ln16.dydx|x=1=34ln2=3ln16.

Checkpoint 3.53

Find the slope for the line tangent to y=3xy=3x at x=2.x=2.

Logarithmic Differentiation

At this point, we can take derivatives of functions of the form y=(g(x))ny=(g(x))n for certain values of n,n, as well as functions of the form y=bg(x),y=bg(x), where b>0b>0 and b1.b1. Unfortunately, we still do not know the derivatives of functions such as y=xxy=xx or y=xπ.y=xπ. These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form h(x)=g(x)f(x).h(x)=g(x)f(x). It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of y=x2x+1exsin3x.y=x2x+1exsin3x. We outline this technique in the following problem-solving strategy.

Problem-Solving Strategy: Using Logarithmic Differentiation
  1. To differentiate y=h(x)y=h(x) using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain lny=ln(h(x)).lny=ln(h(x)).
  2. Use properties of logarithms to expand ln(h(x))ln(h(x)) as much as possible.
  3. Differentiate both sides of the equation. On the left we will have 1ydydx.1ydydx.
  4. Multiply both sides of the equation by yy to solve for dydx.dydx.
  5. Replace yy by h(x).h(x).

Example 3.81

Using Logarithmic Differentiation

Find the derivative of y=(2x4+1)tanx.y=(2x4+1)tanx.

Solution

Use logarithmic differentiation to find this derivative.

lny=ln(2x4+1)tanxStep 1. Take the natural logarithm of both sides.lny=tanxln(2x4+1)Step 2. Expand using properties of logarithms.1ydydx=sec2xln(2x4+1)+8x32x4+1·tanxStep 3. Differentiate both sides. Use theproduct rule on the right.dydx=y·(sec2xln(2x4+1)+8x32x4+1·tanx)Step 4. Multiply byyon both sides.dydx=(2x4+1)tanx(sec2xln(2x4+1)+8x32x4+1·tanx)Step 5. Substitutey=(2x4+1)tanx.lny=ln(2x4+1)tanxStep 1. Take the natural logarithm of both sides.lny=tanxln(2x4+1)Step 2. Expand using properties of logarithms.1ydydx=sec2xln(2x4+1)+8x32x4+1·tanxStep 3. Differentiate both sides. Use theproduct rule on the right.dydx=y·(sec2xln(2x4+1)+8x32x4+1·tanx)Step 4. Multiply byyon both sides.dydx=(2x4+1)tanx(sec2xln(2x4+1)+8x32x4+1·tanx)Step 5. Substitutey=(2x4+1)tanx.

Example 3.82

Using Logarithmic Differentiation

Find the derivative of y=x2x+1exsin3x.y=x2x+1exsin3x.

Solution

This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.

lny=lnx2x+1exsin3xStep 1. Take the natural logarithm of both sides.lny=lnx+12ln(2x+1)xlne3lnsinxStep 2. Expand using properties of logarithms.1ydydx=1x+12x+113cosxsinxStep 3. Differentiate both sides.dydx=y(1x+12x+113cotx)Step 4. Multiply byyon both sides.dydx=x2x+1exsin3x(1x+12x+113cotx)Step 5. Substitutey=x2x+1exsin3x.lny=lnx2x+1exsin3xStep 1. Take the natural logarithm of both sides.lny=lnx+12ln(2x+1)xlne3lnsinxStep 2. Expand using properties of logarithms.1ydydx=1x+12x+113cosxsinxStep 3. Differentiate both sides.dydx=y(1x+12x+113cotx)Step 4. Multiply byyon both sides.dydx=x2x+1exsin3x(1x+12x+113cotx)Step 5. Substitutey=x2x+1exsin3x.

Example 3.83

Extending the Power Rule

Find the derivative of y=xry=xr where rr is an arbitrary real number.

Solution

The process is the same as in Example 3.82, though with fewer complications.

lny=lnxrStep 1. Take the natural logarithm of both sides.lny=rlnxStep 2. Expand using properties of logarithms.1ydydx=r1xStep 3. Differentiate both sides.dydx=yrxStep 4. Multiply byyon both sides.dydx=xrrxStep 5. Substitutey=xr.dydx=rxr1Simplify.lny=lnxrStep 1. Take the natural logarithm of both sides.lny=rlnxStep 2. Expand using properties of logarithms.1ydydx=r1xStep 3. Differentiate both sides.dydx=yrxStep 4. Multiply byyon both sides.dydx=xrrxStep 5. Substitutey=xr.dydx=rxr1Simplify.
Checkpoint 3.54

Use logarithmic differentiation to find the derivative of y=xx.y=xx.

Checkpoint 3.55

Find the derivative of y=(tanx)π.y=(tanx)π.

Section 3.9 Exercises

For the following exercises, find f(x)f(x) for each function.

331.

f(x)=x2exf(x)=x2ex

332.

f(x)=exxf(x)=exx

333.

f(x)=ex3lnxf(x)=ex3lnx

334.

f(x)=e2x+2xf(x)=e2x+2x

335.

f(x)=exexex+exf(x)=exexex+ex

336.

f(x)=10xln10f(x)=10xln10

337.

f(x)=24x+4x2f(x)=24x+4x2

338.

f(x)=3sin3xf(x)=3sin3x

339.

f(x)=xπ·πxf(x)=xπ·πx

340.

f(x)=ln(4x3+x)f(x)=ln(4x3+x)

341.

f(x)=ln5x7f(x)=ln5x7

342.

f(x)=x2ln9xf(x)=x2ln9x

343.

f(x)=log(secx)f(x)=log(secx)

344.

f(x)=log7(6x4+3)5f(x)=log7(6x4+3)5

345.

f(x)=2x·log37x24f(x)=2x·log37x24

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

346.

y=xxy=xx

347.

y=(sin2x)4xy=(sin2x)4x

348.

y=(lnx)lnxy=(lnx)lnx

349.

y=xlog2xy=xlog2x

350.

y=(x21)lnxy=(x21)lnx

351.

y=xcotxy=xcotx

352.

y=x+11x243y=x+11x243

353.

y=x−1/2(x2+3)2/3(3x4)4y=x−1/2(x2+3)2/3(3x4)4

354.

[T] Find an equation of the tangent line to the graph of f(x)=4xe(x21)f(x)=4xe(x21) at the point where

x=−1.x=−1. Graph both the function and the tangent line.

355.

[T] Find the equation of the line that is normal to the graph of f(x)=x·5xf(x)=x·5x at the point where x=1.x=1. Graph both the function and the normal line.

356.

[T] Find the equation of the tangent line to the graph of x3xlny+y3=2x+5x3xlny+y3=2x+5 at the point where x=2.x=2. (Hint: Use implicit differentiation to find dydx.)dydx.) Graph both the curve and the tangent line.

357.

Consider the function y=x1/xy=x1/x for x>0.x>0.

  1. Determine the points on the graph where the tangent line is horizontal.
  2. Determine the points on the graph where y>0y>0 and those where y<0.y<0.
358.

The formula I(t)=sintetI(t)=sintet is the formula for a decaying alternating current.

  1. Complete the following table with the appropriate values.
    tt sintetsintet
    0 (i)
    π2π2 (ii)
    ππ (iii)
    3π23π2 (iv)
    2π2π (v)
    2π2π (vi)
    3π3π (vii)
    7π27π2 (viii)
    4π4π (ix)
  2. Using only the values in the table, determine where the tangent line to the graph of I(t)I(t) is horizontal.
359.

[T] The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year.

  1. Write the exponential function that relates the total population as a function of t.t.
  2. Use a. to determine the rate at which the population is increasing in tt years.
  3. Use b. to determine the rate at which the population is increasing in 10 years.
360.

[T] An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present.

  1. Write the exponential function that relates the amount of substance remaining as a function of t,t, measured in hours.
  2. Use a. to determine the rate at which the substance is decaying in tt hours.
  3. Use b. to determine the rate of decay at t=4t=4 hours.
361.

[T] The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function

N(t)=5.3e0.093t20.87t,(0t4),N(t)=5.3e0.093t20.87t,(0t4), where N(t)N(t) gives the number of cases (in thousands) and t is measured in years, with t=0t=0 corresponding to the beginning of 1960.

  1. Show work that evaluates N(0)N(0) and N(4).N(4). Briefly describe what these values indicate about the disease in New York City.
  2. Show work that evaluates N(0)N(0) and N(3).N(3). Briefly describe what these values indicate about the disease in the United States.
362.

[T] The relative rate of change of a differentiable function y=f(x)y=f(x) is given by 100·f(x)f(x)%.100·f(x)f(x)%. One model for population growth is a Gompertz growth function, given by P(x)=aeb·ecxP(x)=aeb·ecx where a,b,a,b, and cc are constants.

  1. Find the relative rate of change formula for the generic Gompertz function.
  2. Use a. to find the relative rate of change of a population in x=20x=20 months when a=204,b=0.0198,a=204,b=0.0198, and c=0.15.c=0.15.
  3. Briefly interpret what the result of b. means.

For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.

Years since 1790 Population
0 33,131
10 60,515
20 96,373
30 123,706
40 202,300
50 312,710
60 515,547
70 813,669
Table 3.8 New York City Population Over Time Source: http://en.wikipedia.org/wiki/Largest_cities_in_the_United_States
_by_population_by_decade.
363.

[T] Using a computer program or a calculator, fit a growth curve to the data of the form p=abt.p=abt.

364.

[T] Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.

365.

[T] Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.

366.

[T] Using the tables of first and second derivatives and the best fit, answer the following questions:

  1. Will the model be accurate in predicting the future population of New York City? Why or why not?
  2. Estimate the population in 2010. Was the prediction correct from a.?
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