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Chapter 5 Solutions

Worked solutions to all thirty problems, with the reason for each step. These explanations were added for this edition; the chapter answer key and original worked examples follow below.

Chapter problems → · Chapter answer key → · Original worked examples →

Problem 1

Original problem · clarified · added worked solution

You deposit $1000 into the stock market that you plan to keep until you retire in 45 years. Your portfolio has an annual rate of return of 10% per year. If you are 20 years old, at what rate is your wealth growing when you are 60? (Use units of dollars per year)

Clarification: Assume continuous compounding at an annual rate of 0.10, with t in years since the deposit.

Authored worked explanation for the original source problem.

Source review note: The source answer matches continuous compounding. “Annual rate” alone does not specify continuous versus discrete compounding; this reviewed explanation states the implied convention.

Use the source's continuous model $A (t) = 1000 e^{0.1 t}$ , with t measured from age 20.
Why: The source answer implies this model; annual discrete compounding would be a different model.
At age 60, t=40, and $A' (t) = 100 e^{0.1 t}$ gives $A' (40) = 100 e^{4}$ .
Why: Differentiate with respect to years; age 65 and the full 45 years are not the requested instant.

Reviewed answer: $100 e^{4} \approx 5460$ dollars/year under continuous compounding.

Takeaway: State the growth model and identify elapsed time before differentiating.

Final answer

100 e^{4} \approx 5460

dollars/year under continuous compounding.

Return to problem 1 →

Problem 2

Original problem · clarified · added worked solution

Find

lim_{x \to 3} \frac{x - 3}{\sqrt{x - 3}} + x \cdot \ln (x - 3) - 3 \ln (x - 3)

Clarification: Take the real-domain limit as x approaches 3 from the right.

Authored worked explanation for the original source problem.

Source review note: The source result 0 is consistent for x→3⁺. The printed function has no real left-side domain near 3.

Set h=x−3>0; the expression becomes $\sqrt{h} + h \ln h$ .
Why: The log and square root require approaching 3 through x>3.
$\sqrt{h} \to 0$ . For the other term, write $h \ln h = \ln h / (1 / h)$ .
Why: This converts 0·(−∞) into a quotient to which L'Hospital applies.
Its derivative ratio is $(1 / h) / (- 1 / h^{2}) = - h \to 0$ , so the sum tends to 0.
Why: The numerator and denominator are differentiable for h>0.

Reviewed answer: $0$ from the right.

Takeaway: A domain can force a one-sided limit even when the prompt omits the side.

Final answer

0

from the right.

Return to problem 2 →

Problem 3

Original problem · added worked solution

Solve

lim_{x \to \infty} x^{- x^{x}}

Authored worked explanation for the original source problem.

For x>1, write $x^{- x^{x}} = e^{- x^{x} \ln x}$ .
Why: Taking a logarithm turns a variable power into a product in the exponent.
Both x^x and ln x grow without bound, so the exponent tends to −∞ and the exponential tends to 0.
Why: The factors are positive, so their product cannot cancel.

Reviewed answer: $0$

Takeaway: A negative exponent growing in magnitude forces exponential decay.

Final answer

0

Return to problem 3 →

Problem 4

Original problem · added worked solution

What is

lim_{x \to \infty} \sin (\frac{\ln x}{x})

Authored worked explanation for the original source problem.

$\ln x / x$ is ∞/∞ as x→∞, so L'Hospital gives $(1 / x) / 1 \to 0$ .
Why: The differentiated denominator stays nonzero.
Sine is continuous at 0, so the outside sine tends to sin0=0.
Why: Carry the inner limit through the continuous outer function.

Reviewed answer: $0$

Takeaway: Solve the inner limit before applying the trig function.

Final answer

0

Return to problem 4 →

Problem 5

Original problem · added worked solution

Calculate the derivative of

f (x) = \log_{4} (x^{2}) + 3^{x \cdot \log_{2} (x)} + 4 \ln (x)

Authored worked explanation for the original source problem.

Source review note: The source uses ln2 as the multiplier for differentiating 3^u; the correct multiplier is ln3.

$d [\log_{4} (x^{2})] / d x = 2 / (x \ln 4)$ , and $d (4 \ln x) / d x = 4 / x$ .
Why: Use change of base or the general log derivative, with the original x>0 domain.
For the middle term, let u=x log₂x, so $u' = \log_{2} x + 1 / \ln 2$ .
Why: The exponent is itself a product.
$d (3^{u}) / d x = 3^{u} \ln 3 \cdot u'$ .
Why: The logarithmic multiplier belongs to the exponential's base 3.

Reviewed answer: $2 / (x \ln 4) + 3^{x \log_{2} x} \ln 3 \cdot (\log_{2} x + 1 / \ln 2) + 4 / x$ , x>0.

Takeaway: Keep the exponential base and the logarithm base separate.

Final answer

2 / (x \ln 4) + 3^{x \log_{2} x} \ln 3 \cdot (\log_{2} x + 1 / \ln 2) + 4 / x

, x>0.

Return to problem 5 →

Problem 6

Original problem · added worked solution

Find the derivative of

f (x) = \tan^{- 1} (\log_{2} x) + 3 \sin^{- 1} (\cos^{- 1} x)

Authored worked explanation for the original source problem.

The arctangent term gives $(1 / (1 + (\log_{2} x)^{2})) (1 / (x \ln 2))$ .
Why: Apply the inverse tangent derivative and the inside logarithm derivative.
The arcsine term gives $3 (1 / \sqrt{1 - (\arccos x)^{2}}) (- 1 / \sqrt{1 - x^{2}})$ .
Why: The arccos derivative supplies the negative sign.
Require $\arccos x \in [0, 1]$ , so the function domain is [cos1,1]; use the interior for the displayed finite derivative.
Why: A composite inverse trig function needs both inside and outside domain checks.

Reviewed answer: $1 / [x \ln 2 (1 + (\log_{2} x)^{2})] - 3 / [\sqrt{1 - (\arccos x)^{2}} \sqrt{1 - x^{2}}]$ , for cos1<x<1.

Takeaway: The outer inverse trig domain restricts the inner inverse trig output.

Final answer

1 / [x \ln 2 (1 + (\log_{2} x)^{2})] - 3 / [\sqrt{1 - (\arccos x)^{2}} \sqrt{1 - x^{2}}]

, for cos1<x<1.

Return to problem 6 →

Problem 7

Original problem · added worked solution

Find

lim_{x \to \infty} {(4 + \frac{3}{x})}^{\frac{1}{x}}

Authored worked explanation for the original source problem.

Take logarithms: $\ln y = \ln (4 + 3 / x) / x$ .
Why: The base is positive near positive infinity.
The numerator tends to ln4 while the denominator grows without bound, so ln y→0.
Why: This quotient has a finite numerator rather than an indeterminate ∞/∞ form.
Then y→e⁰=1.
Why: Exponentiation is continuous.

Reviewed answer: $1$

Takeaway: A shrinking exponent can send a fixed positive base toward 1.

Final answer

1

Return to problem 7 →

Problem 8

Original problem · added worked solution

Find the derivative of

f (x) = 5 \sec^{- 1} (4^{x}) + 2 \cot^{- 1} (\csc^{- 1} (x)) + 8^{x^{\ln x}}

Authored worked explanation for the original source problem.

Source review note: The source omits the coefficient 5, uses plus signs in inverse secant/cosecant radicals that require minus signs, and does not correctly differentiate x^(ln x).

For arcsec(4^x), u=4^x>0 and $d (arcsec u) / d x = u' / (| u | \sqrt{u^{2} - 1})$ , giving $\ln 4 / \sqrt{4^{2 x} - 1}$ before the factor 5.
Why: Arcsec uses u²−1 and an absolute value, not u²+1.
For arccot(arccsc x), both inside and outside derivatives are negative; on x>1 their product gives $1 / [x \sqrt{x^{2} - 1} (1 + (arccsc x)^{2})]$ before the factor 2.
Why: Use the principal arccot convention with derivative −1/(1+u²).
For v=x^(ln x), $\ln v = (\ln x)^{2}$ , so $v' = 2 \ln x \cdot x^{\ln x - 1}$ .
Why: Logarithmic differentiation handles both variable base and exponent.
The last term differentiates to $8^{v} \ln 8 \cdot v'$ ; the combined function's real interior domain is x>1.
Why: The logarithmic power requires x>0, while arccsc requires |x|≥1.

Reviewed answer: $5 \ln 4 / \sqrt{4^{2 x} - 1} + 2 / [x \sqrt{x^{2} - 1} (1 + (arccsc x)^{2})] + 8^{x^{\ln x}} \ln 8 \cdot 2 \ln x \cdot x^{\ln x - 1}$ , for x>1.

Takeaway: State inverse-function conventions and check the combined domain.

Final answer

5 \ln 4 / \sqrt{4^{2 x} - 1} + 2 / [x \sqrt{x^{2} - 1} (1 + (arccsc x)^{2})] + 8^{x^{\ln x}} \ln 8 \cdot 2 \ln x \cdot x^{\ln x - 1}

, for x>1.

Return to problem 8 →

Problem 9

Original problem · added worked solution

Find the slope of the tangent line at x=0 of

f (x) = 2 \sin^{- 1} (x) + 3^{x}

Authored worked explanation for the original source problem.

$f' (x) = 2 / \sqrt{1 - x^{2}} + 3^{x} \ln 3$ .
Why: The inverse sine and base-3 exponential use different special rules.
At x=0 this is 2+ln3.
Why: Both terms are defined and finite at zero.

Reviewed answer: $2 + \ln 3$

Takeaway: Evaluate the complete derivative at the requested point.

Final answer

2 + \ln 3

Return to problem 9 →

Problem 10

Original problem · added worked solution

Find the slope of the tangent line at x=0.5 of

f (x) = 3 \sec^{- 1} (4^{\log_{2} x}) + 2 \cos^{- 1} (\sin^{- 1} (x)) + 2^{x \cdot \ln (3 x)}

Authored worked explanation for the original source problem.

Source review note: The source gives 4.31, but arcsec(4^(log₂(0.5)))=arcsec(1/4) is not real-defined.

$4^{\log_{2} x} = x^{2}$ for x>0, so at x=0.5 the arcsec input is 1/4.
Why: Rewrite the exponential using 4=2².
Real arcsec requires its input to have absolute value at least 1.
Why: A real secant value cannot lie strictly between −1 and 1.
The function is therefore not real-defined at the requested input, so it has no real tangent slope there.
Why: Differentiation cannot repair a domain failure.

Reviewed answer: No real tangent slope at x=0.5.

Takeaway: Check a nested inverse trig input before computing its derivative.

Final answer

No real tangent slope at x=0.5.

Return to problem 10 →

Problem 11

Added problem and worked solution

Differentiate $f (x) = \ln (x^{2} + 1)$ .

Authored solution to an added practice problem.

The derivative of ln u is u′/u.
Why: The argument x²+1 is positive for all real x.
With u′=2x, the result is 2x/(x²+1).
Why: The chain rule multiplies the outer derivative by the inner derivative.

Reviewed answer: $2 x / (x^{2} + 1)$

Takeaway: A log derivative retains its argument in the denominator.

Final answer

2 x / (x^{2} + 1)

Return to problem 11 →

Problem 12

Added problem and worked solution

Find the derivative of $f (x) = 5^{1 - 2 x}$ .

Authored solution to an added practice problem.

The derivative of 5^u is 5^u ln5·u′.
Why: The base determines the logarithmic multiplier.
For u=1−2x, u′=−2.
Why: Include the exponent's slope.

Reviewed answer: $- 2 \cdot 5^{1 - 2 x} \ln 5$

Takeaway: A decreasing exponent gives a negative derivative here.

Final answer

- 2 \cdot 5^{1 - 2 x} \ln 5

Return to problem 12 →

Problem 13

Added problem and worked solution

Find the derivative and domain of $f (x) = \log_{3} (x^{2} - 4)$ .

Authored solution to an added practice problem.

The original logarithm requires x²−4>0.
Why: A real log has a positive argument.
Differentiate by u′/(u ln3) with u=x²−4.
Why: The general log's base 3 contributes ln3 in the denominator.

Reviewed answer: $2 x / [(x^{2} - 4) \ln 3]$ , domain $(- \infty, - 2) \cup (2, \infty)$ .

Takeaway: Solve the argument inequality before using a log derivative.

Final answer

2 x / [(x^{2} - 4) \ln 3]

, domain

(- \infty, - 2) \cup (2, \infty)

Return to problem 13 →

Problem 14

Added problem and worked solution

Use logarithmic differentiation to differentiate $y = x^{x}$ for $x > 0$ .

Authored solution to an added practice problem.

Set y=x^x and take logs: $\ln y = x \ln x$ .
Why: This removes the variable exponent.
Differentiate to $y' / y = \ln x + 1$ , then multiply by y.
Why: The right side uses the product rule.

Reviewed answer: $x^{x} (\ln x + 1)$ , x>0.

Takeaway: A variable base and exponent both contribute.

Final answer

x^{x} (\ln x + 1)

, x>0.

Return to problem 14 →

Problem 15

Added problem and worked solution

Differentiate $y = (x^{2} + 1)^{x}$ .

Authored solution to an added practice problem.

Let y=(x²+1)^x, so $\ln y = x \ln (x^{2} + 1)$ .
Why: The base is positive for every real x.
Then $y' / y = \ln (x^{2} + 1) + x \cdot 2 x / (x^{2} + 1)$ , and multiply by y.
Why: Product and chain rules account for the two changing parts.

Reviewed answer: $(x^{2} + 1)^{x} [\ln (x^{2} + 1) + 2 x^{2} / (x^{2} + 1)]$

Takeaway: Logarithms turn a difficult power into a manageable product.

Final answer

(x^{2} + 1)^{x} [\ln (x^{2} + 1) + 2 x^{2} / (x^{2} + 1)]

Return to problem 15 →

Problem 16

Added problem and worked solution

Differentiate $f (x) = \arcsin (3 x)$ and give the finite derivative domain.

Authored solution to an added practice problem.

Use the arcsine derivative 1/√(1−u²).
Why: The finite derivative requires the argument strictly inside [−1,1].
The inside 3x has derivative 3, giving the answer.
Why: The function itself includes endpoints, but its derivative blows up there.

Reviewed answer: $3 / \sqrt{1 - 9 x^{2}}$ , for |x|<1/3.

Takeaway: Differentiate on the interior even when the original function includes endpoints.

Final answer

3 / \sqrt{1 - 9 x^{2}}

, for |x|<1/3.

Return to problem 16 →

Problem 17

Added problem and worked solution

Differentiate $f (x) = \arccos (x^{2})$ on its interior domain.

Authored solution to an added practice problem.

Set u=x², whose derivative is 2x.
Why: The outer arccos requires 0≤x²≤1.
The inverse cosine derivative is −u′/√(1−u²), giving the answer.
Why: Keep the outer negative sign and square the complete inner expression.

Reviewed answer: $- 2 x / \sqrt{1 - x^{4}}$ for |x|<1.

Takeaway: A squared input becomes a fourth power in the radical.

Final answer

- 2 x / \sqrt{1 - x^{4}}

for |x|<1.

Return to problem 17 →

Problem 18

Added problem and worked solution

Find the derivative of $f (x) = \arctan (e^{x})$ .

Authored solution to an added practice problem.

Use the arctangent derivative u′/(1+u²).
Why: Arctangent accepts every real inner value.
For u=e^x, u′=e^x and u²=e^(2x).
Why: The exponential's square doubles its exponent.

Reviewed answer: $e^{x} / (1 + e^{2 x})$

Takeaway: An inverse tangent composition has a plus sign in its denominator.

Final answer

e^{x} / (1 + e^{2 x})

Return to problem 18 →

Problem 19

Added problem and worked solution

For the one-to-one function $f (x) = x^{3} + x$ , find $(f^{- 1})' (10)$ without solving for the full inverse.

Authored solution to an added practice problem.

Since f(2)=8+2=10, the inverse sends 10 to 2.
Why: Locate the original input corresponding to the inverse input.
$f' (2) = 3 \cdot 2^{2} + 1 = 13$ , so the inverse slope is its reciprocal, 1/13.
Why: The inverse derivative rule requires a nonzero original slope.

Reviewed answer: $(f^{- 1})' (10) = 1 / 13$ .

Takeaway: Inverse slopes are reciprocal at corresponding points.

Final answer

(f^{- 1})' (10) = 1 / 13

Return to problem 19 →

Problem 20

Added problem and worked solution

A locally invertible function has $f (3) = 7$ and $f' (3) = - 2$ . Find the derivative of its inverse at 7.

Authored solution to an added practice problem.

For g=f⁻¹, $f (3) = 7$ means g(7)=3.
Why: Match the two functions' corresponding coordinates.
$g' (7) = 1 / f' (3) = 1 / (- 2) = - 1 / 2$ .
Why: The given derivative is nonzero, so the local inverse rule applies.

Reviewed answer: $g' (7) = - 1 / 2$ .

Takeaway: A negative inverse slope stays negative.

Final answer

g' (7) = - 1 / 2

Return to problem 20 →

Problem 21

Added problem and worked solution

Use L’Hospital’s rule to evaluate $\lim_{x \to 0} (e^{2 x} - 1) / x$ .

Authored solution to an added practice problem.

Both numerator and denominator approach 0.
Why: This is an eligible indeterminate quotient.
Differentiate them separately to get $2 e^{2 x} / 1$ , whose limit is 2.
Why: L'Hospital differentiates the quotient's two parts, not the quotient as a whole.

Reviewed answer: $2$

Takeaway: Check the form before applying the rule.

Final answer

2

Return to problem 21 →

Problem 22

Added problem and worked solution

Evaluate $\lim_{x \to 0} (e^{x} - 1 - x) / x^{2}$ .

Authored solution to an added practice problem.

The first derivative ratio is $(e^{x} - 1) / (2 x)$ , still 0/0.
Why: The original numerator and its first derivative both vanish at zero.
Differentiate again to $e^{x} / 2$ , giving 1/2.
Why: Reapply the rule only after checking the new indeterminate form.

Reviewed answer: $1 / 2$

Takeaway: More than one derivative may be needed.

Final answer

1 / 2

Return to problem 22 →

Problem 23

Added problem and worked solution

Find $\lim_{x \to 0^{+}} x \ln x$ .

Authored solution to an added practice problem.

Write $x \ln x = \ln x / (1 / x)$ for x>0.
Why: A product 0·(−∞) must be converted to an eligible quotient.
The derivative ratio is −x, which tends to 0 as x→0⁺.
Why: Both numerator and denominator are differentiable on the right-hand domain.

Reviewed answer: $0$

Takeaway: Rewrite a product before reaching for L'Hospital.

Final answer

0

Return to problem 23 →

Problem 24

Added problem and worked solution

Evaluate $\lim_{x \to \infty} x^{2} / e^{x}$ .

Authored solution to an added practice problem.

The quotient x²/e^x is ∞/∞.
Why: Both parts grow without bound.
Apply L'Hospital twice: $2 x / e^{x}$ , then $2 / e^{x} \to 0$ .
Why: Each differentiation lowers the polynomial degree while leaving the exponential.

Reviewed answer: $0$

Takeaway: Exponentials dominate any fixed polynomial power here.

Final answer

0

Return to problem 24 →

Problem 25

Added problem and worked solution

Evaluate $\lim_{x \to 0} (1 + 2 x)^{1 / x}$ .

Authored solution to an added practice problem.

Set y=(1+2x)^(1/x) for x near zero and take $\ln y = \ln (1 + 2 x) / x$ .
Why: The positive base permits logarithms on both sides near zero.
L'Hospital gives the logarithmic limit $[2 / (1 + 2 x)] / 1 \to 2$ .
Why: The log quotient is 0/0.
Thus y→e².
Why: Exponentiate the logarithmic limit.

Reviewed answer: $e^{2}$

Takeaway: Take logs to turn a variable power into a quotient.

Final answer

e^{2}

Return to problem 25 →

Problem 26

Added problem and worked solution

A student uses L’Hospital on $\lim_{x \to 0} (e^{x} + 1) / (x + 1)$ and gets 1. Diagnose the mistake.

Authored solution to an added practice problem.

At zero, the numerator e^x+1 tends to 2 and denominator x+1 to 1.
Why: The form is 2/1, not 0/0 or ∞/∞.
Direct substitution gives 2; differentiating numerator and denominator would give 1 and an unjustified result.
Why: The rule's hypotheses are essential.

Reviewed answer: The limit is 2 by substitution; L’Hospital cannot be used on the original quotient.

Takeaway: A recognizable quotient is not automatically an indeterminate quotient.

Final answer

The limit is 2 by substitution; L’Hospital cannot be used on the original quotient.

Return to problem 26 →

Problem 27

Added problem and worked solution

In a fictional savings model, $A (t) = 2000 e^{0.04 t}$ dollars, with t in years. Find the instantaneous growth rate at year 5.

Authored solution to an added practice problem.

Differentiate the stipulated balance $A (t) = 2000 e^{0.04 t}$ to $A' = 80 e^{0.04 t}$ .
Why: The exponent coefficient has units of inverse years.
At t=5 substitute to obtain 80e^0.2.
Why: This is an instantaneous rate, not the amount earned during a whole year.

Reviewed answer: $A' (5) = 80 e^{0.2} \approx 97.71$ dollars/year.

Takeaway: Units distinguish a balance from its growth rate.

Final answer

A' (5) = 80 e^{0.2} \approx 97.71

dollars/year.

Return to problem 27 →

Problem 28

Added problem and worked solution

A fictional substance follows $M (t) = 10 e^{- k t}$ grams and halves in 6 hours. Find k and its mass rate at hour 6.

Authored solution to an added practice problem.

The equation $10 e^{- 6 k} = 5$ gives $e^{- 6 k} = 1 / 2$ , so k=ln2/6.
Why: A half-life specifies when the mass is halved.
$M' = - k M$ , so at hour 6, where M=5, the rate is −5k.
Why: The negative sign indicates decreasing mass.

Reviewed answer: $k = \ln 2 / 6$ h⁻¹; $M' (6) = - (5 \ln 2) / 6$ grams/hour.

Takeaway: Use the current mass after finding the rate constant.

Final answer

k = \ln 2 / 6

h⁻¹;

M' (6) = - (5 \ln 2) / 6

grams/hour.

Return to problem 28 →

Problem 29

Added problem and worked solution

State the derivative of $arcsec x$ for $| x | > 1$ and explain why $| x |$ is needed.

Authored solution to an added practice problem.

From $\sec y = x$ , implicit differentiation gives $\sec y \tan y \cdot y' = 1$ .
Why: Choose the principal arcsec branch.
The branch sign gives $\sec y \tan y = | x | \sqrt{x^{2} - 1}$ , so solve for y′.
Why: The absolute value makes the correct derivative positive on both allowed intervals.

Reviewed answer: $1 / (| x | \sqrt{x^{2} - 1})$ , for |x|>1.

Takeaway: The inverse secant formula needs an absolute value.

Final answer

1 / (| x | \sqrt{x^{2} - 1})

, for |x|>1.

Return to problem 29 →

Problem 30

Added problem and worked solution

Show that the derivative of $\ln | x |$ is $1 / x$ on both x>0 and x<0.

Authored solution to an added practice problem.

For x>0, ln|x|=ln x and its derivative is 1/x.
Why: Use the usual log derivative.
For x<0, ln|x|=ln(−x), whose derivative is $(1 / (- x)) (- 1) = 1 / x$ .
Why: The inside minus sign cancels the denominator's minus sign.

Reviewed answer: $1 / x$ , for x≠0.

Takeaway: The absolute-value log extends the primitive to each nonzero sign interval.

Final answer

1 / x

, for x≠0.

Return to problem 30 →

Chapter answer key and worked steps

Answers and worked steps for the original problems, including verified corrections where needed.

Checked answer

$100 e^{4} \approx 5460$ dollars/year under continuous compounding.
Problem 1 →
Checked answer

$0$ from the right.
Problem 2 →
Original short answer

0
Problem 3 →
Original short answer

0
Problem 4 →
Corrected answer

$2 / (x \ln 4) + 3^{x \log_{2} x} \ln 3 \cdot (\log_{2} x + 1 / \ln 2) + 4 / x$ , x>0.
Problem 5 →
Original short answer

$f^{'} (x) = \frac{1}{x \cdot \ln 2 \cdot (1 + (\log_{2} x)^{2})} - \frac{3}{\sqrt{(1 - \cos^{- 1} (x)^{2}) \cdot (1 - x^{2})}}$
Problem 6 →
Original short answer

$1$
Problem 7 →
Corrected answer

$5 \ln 4 / \sqrt{4^{2 x} - 1} + 2 / [x \sqrt{x^{2} - 1} (1 + (arccsc x)^{2})] + 8^{x^{\ln x}} \ln 8 \cdot 2 \ln x \cdot x^{\ln x - 1}$ , for x>1.
Problem 8 →
Original short answer

$f^{'} (0) = 2 + \ln 3$
Problem 9 →
Corrected answer

No real tangent slope at x=0.5.
Problem 10 →

Read the chapter answer key in the book →

Source worked examples

These are the book’s example problems and worked steps, with the original explanations.

Example Problem 5.1.1

Powers and Logarithms →

Differentiate $y = \frac{x \cdot \log_{2} (x) + 3^{\sin x}}{y}$ and set the derivative as a function of x only.

Let’s multiply both sides by y first so that we won’t have to take the quotient rule:

$y^{2} = x \cdot \log_{2} x + 3^{\sin x}$
Now we can take the derivative of both sides
$\frac{d}{d x} (y^{2} = x \cdot \log_{2} x + 3^{\sin x})$
$2 y \cdot \frac{d y}{d x} = 1 \cdot \log_{2} (x) + x \cdot \frac{1}{x \cdot \ln 2} + 3^{\sin x} \cdot \ln 3 \cdot \cos x$
Since we are getting into complex derivatives, let’s create a key for each rule.

$Magenta$ - Quotient Rule
$Green$ - Chain Rule
$Blue$ - Product Rule

$\frac{d y}{d x} = \frac{\log_{2} x + x \cdot \frac{1}{x \cdot \ln 2} + 3^{\sin x} \cdot \ln 3 \cdot \cos x}{2 y}$

We need to figure out what y is equal to. Since
$y^{2} = x \log_{2} x + 3^{\sin x}$ ,

$y = \sqrt{x \log_{2} (x) + 3^{\sin x}}$
For simplicity’s sake, we’ll only be concerned with the positive square root. Substituting this equation into the derivative gives:

$\frac{d y}{d x} = \frac{\log_{2} x + \frac{1}{\ln 2} + 3^{\sin x} \cdot \ln 3 \cdot \cos x}{2 \sqrt{x \cdot \log_{2} (x) + 3^{\sin x}}}$

Gallery 5.1.1 Example Problem 5.1.1

Read all original image captions

Image 1

The graph of the function is shown in blue, and the graph of the derivative function is shown in green. Let’s view the x-axis on a logarithmic scale to see if we can pick up any patterns, because this does not tell us very much

Image 2

Using a logarithmic x-axis works especially well in this case because our domain is x>0, which also brings us to a disadvantage of logarithmic graphs. They cannot show negative values! Notice the value of each notch on the x-axis. Would using a logarithmic y-axis be a good idea?

Image 3

In this case, most definitely not. Before, the derivative crossed y=0 when the slope of the tangent line was 0 (see orange and red dotted lines). Now, since log(0) is undefined, the derivative graph approaches -∞ instead. Not only that, we cannot see when the slope is negative!!

Return to the original example →

Example Problem 5.2.1

Inverse Functions →

Find the derivative of the inverse of $f (x) = x^{2} + 2 x + 1$ using (a) the formula given above, and (b) by calculating the inverse first, then taking its derivative. Which method is easier?

Next is a table summarizing the trigonometric inverse functions and their derivatives.

In order to use the formula, we must calculate the derivative of the original function as well as the inverse of the original function. As a reminder, the formula is:

$\frac{d}{d x} f^{- 1} (x) = \frac{1}{f^{'} (f^{- 1} (x))}$

We’ll start by creating a table and filling it in as we go.

$\begin{aligned} f (x) & = x^{2} + 2 x + 1 & \frac{d}{d x} f (x) & = 2 x + 2 \\ f^{- 1} (x) & = ? & \frac{d}{d x} f^{- 1} (x) & = ? \end{aligned}$

To calculate $y = f^{- 1} (x)$ , we simply switch the x and y in the original equation, which gives us $x = y^{2} + 2 y + 1$ , where $y = f^{- 1} (x)$

$x = (y + 1)^{2}$

$\sqrt{x} = y + 1$

$f^{- 1} (x) = \sqrt{x} - 1$

For the next step, we simply plug this into the formula, but it’s tricky, so watch carefully!

$\frac{d}{d x} f^{- 1} (x) = \frac{1}{f^{'} (f^{- 1} (x))} = \frac{1}{2 (\sqrt{x} - 1) + 2}$

Simplifying this gives us $\frac{d}{d x} f^{- 1} (x) = \frac{1}{2 \sqrt{x}}$

Next, we try without the formula. We know that $f^{- 1} (x) = \sqrt{x} - 1$ from part a). Now we just take the derivative of this normally. The derivative of the constant (–1) is 0, and the derivative of $\sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$ , therefore $\frac{d}{d x} f^{- 1} (x) = \frac{1}{2 \sqrt{x}}$ . We get the same answer!

Source figure

Interactive 5.2.1 Inverse and normal functions.

Magnification 1×

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Two more points

Notice that this pointer goes through the point (0, 1) and (1, 0). The tangent lines at both of these points have a slope of 1. However, since the inverse of 1 is also 1, this does not tell us much.

Second point of interest

Here, we look at the point y=1 (compared to x=1) and notice that the slope of the tangent line is 1/e. Seem familiar? All the derivatives of inverse functions we proved were in the form of 1/(some function). Now it (almost) makes sense!

First point of interest

The slope of the tangent line at x=1 is e, which makes sense since the derivative of e^x is e^x.

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Example Problem 5.2.2

Inverse Functions →

Calculate the second derivative of $y = \tan^{- 1} (\sin x)$ at $x = \frac{π}{2}$

Using the color coding scheme again:

$Magenta$ - Quotient Rule
$Green$ - Chain Rule
$Blue$ - Product Rule

Let’s take the first derivative:

$y^{'} = \frac{1}{1 + \sin^{2} x} \cdot \cos x = \frac{\cos x}{1 + \sin^{2} x}$

$y^{''} = \frac{(1 + \sin^{2} x) \cdot \frac{d}{d x} (\cos x) - \cos x \cdot \frac{d}{d x} (1 + \sin^{2} x)}{(1 + \sin^{2} x)^{2}}$

$y^{''} = \frac{- \sin x (1 + \sin^{2} x) - \cos x \cdot 2 \sin x \cos x}{(1 + \sin^{2} x)^{2}}$

At this point we can just go ahead and plug in $x = \frac{π}{2}$ into the derivative to get the answer (we do not need to simplify since we are looking for a numerical answer). We know that sin(π/2)=1 and cos(π/2)=0 from our unit circle. That’s all we need to solve the rest of this problem.

$a t \frac{π}{2}, y^{''} = \frac{- 1 \cdot (1 + 1^{2}) - 0 \cdot 2 \cdot 1 \cdot 0}{(1 + 1^{2})^{2}} = - \frac{1}{2}$

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Example Problem 5.3.1

L’Hospital’s Rule →

Find $lim_{x \to \infty} \frac{x^{5}}{e^{x}}$

The indeterminate form we encounter if we try to take this limit is ∞/∞. So let’s use L’Hospital’s Rule. We get
$lim_{x \to \infty} \frac{\frac{d}{d x} x^{5}}{\frac{d}{d x} e^{x}} = lim_{x \to \infty} \frac{5 x^{4}}{e^{x}}$
However, this is another indeterminate form (∞/∞), so let’s take the derivative again. This gives us:

$lim_{x \to \infty} \frac{20 x^{3}}{e^{x}}$ You can see that we will keep getting indeterminate forms and keep using L’Hospital’s Rule until the power of x reduces to 0. This gives us the result

$lim_{x \to \infty} \frac{5!}{e^{x}} = lim_{x \to \infty} \frac{120}{e^{x}} = 0$

The limit here is 0 because as x approaches ∞, the denominator approaches ∞, and any constant divided by ∞ is 0.

For a nonnegative integer n, the quotient xⁿ/eˣ has form ∞/∞ as x → ∞ when n > 0. Repeated applications of L’Hospital’s Rule, with its hypotheses checked, reduce the numerator to the constant n! while the exponential denominator remains eˣ. The resulting limit is 0. This finite differentiation argument proves the result for nonnegative integer powers, not by itself for every real exponent.

$lim_{x \to \infty} \frac{x^{n}}{e^{x}} = 0$ for each fixed real n. To extend the integer result, choose a nonnegative integer m greater than n. For x ≥ 1, 0 < xⁿ/eˣ ≤ xᵐ/eˣ. The upper bound tends to 0 by the integer argument, so the squeeze theorem proves the real-power result.

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Example Problem 5.3.2

L’Hospital’s Rule →

Find $lim_{x \to 0^{+}} x^{x}$ Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

The trickiest part here is rewriting the limit as
$lim_{x \to 0} e^{\ln x^{x}} = lim_{x \to 0} e^{x \cdot \ln x}$

Now we’re stuck. We don’t know what the exponent equals as it approaches 0, so let’s prove its limit separately to figure it out.

$lim_{x \to 0} x \ln x = lim_{x \to 0} \frac{\ln x}{\frac{1}{x}}$ (we can rewrite it this way for the same reason that 2*4=2/(1/4), just replace 4 with x, and 2 with ln(x). Now we can take L’Hospital’s Rule for the above limit.

$lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{x}} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} \ln x}{\frac{d}{d x} (\frac{1}{x})} = lim_{x \to 0^{+}} \frac{\frac{1}{x}}{- \frac{1}{x^{2}}} = lim_{x \to 0^{+}} (- x) = 0$

$lim_{x \to 0} x \cdot \ln x = 0$

Let’s plug this back into the original limit.

We already found that x ln(x) → 0 as x → 0⁺. Now use continuity of the exponential function: e^{x ln(x)} → e⁰ = 1. L’Hospital’s Rule was used to evaluate the exponent’s limit; this final step uses continuity.

$lim_{x \to 0^{+}} e^{x \cdot \ln x} = e^{{lim}_{x \to 0^{+}} x \cdot \ln x} = e^{0} = 1$
Therefore, Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

$lim_{x \to 0^{+}} x^{x} = 1$ Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

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Example Problem 5.4.1

FFR - Applications →

Let’s use an idealized carbon-14 decay model with a half-life of about 5700 years. The independently supplied initial amount is B = 4 g, and the present amount is M = 10 mg = 0.01 g. Assume no isotope enters or leaves the sample after the model begins, apart from radioactive decay. These supplied model amounts replace the earlier unsupported calculation from a mammoth’s total body mass. Real radiocarbon dating uses the remaining isotope ratio in once-living material and calibration; we cannot infer an initial carbon-14 amount from total fossil mass alone.

a) According to this decay model, how much time has elapsed?

b) What is the current rate of change of the tracked carbon-14 mass, in mg per millennium?

a) This seems like a really long question, but we can break this down piece by piece. First of all, let’s define half-life. It’s the time it takes for half the mass (or concentration in chemistry) to be consumed. Half life stays constant for first order exponential decay.

The initial amount B = 4 g is a supplied model parameter. It does not follow from the total body mass of an animal. For a dating model, the initial isotope amount or ratio must be established independently before the remaining fraction can determine elapsed time.

After 1 half-life cycle, this mass will become 2g, then after another 5700 years, it’ll become 1g, then 0.5g after another 5700 years, and so on. That’s what the formula M=Be^-rt represents. B represents initial mass when the decay started to occur. M represents present mass. t represents the amount of time that has passed, and r is the rate constant.

As the name implies, r stays constant, but we do not know what it is. How can we figure it out?

We know the definition of a half life, so let’s convert that into equation form:

$1 = 2 e^{- r \cdot 5.7}$ (I used 5.7 here because keeping the units in millennia makes it easier to calculate than keeping them in years. r will just have different units, but it will still get the job done).

Since B represents the initial mass, we just gave it an arbitrary number (2 is the easiest) so that the final mass will be half of B (in this case 1). Now let’s do some algebraic manipulation that gives us
$\ln \frac{1}{2} = \ln e^{- r \cdot 5.7}$
$- 0.693 = - r \cdot 5.7$
$r = 0.1216 {millenia}^{- 1}$

It’s important to keep track of units in problems like these. Had we chosen to go with years instead of millennia, r would be a much smaller value (1000 times smaller) with different units (years^-1). A general rule of thumb is that the exponent should have no units (units should cancel out).

Now use consistent units: B = 4 g, M = 10 mg = 0.01 g, and t is in millennia. The half-life gives r = ln(2)/5.7 ≈ 0.1216 per millennium. Substitute those values into M = B e^−rt.

$0.01 = 4 \cdot e^{- 0.1216 \cdot t}$

$\ln (0.0025) = \ln (e^{- 0.1216 \cdot t})$

Taking logarithms gives ln(0.0025) = −rt, so t = ln(400)/r. Keep the unrounded logarithm until the final calculation.

Using r = ln(2)/5.7, t = 5.7 ln(400)/ln(2) ≈ 49.2700 millennia, or about 49,270 years. With the displayed rounded constant 0.1216, the answer is about 49.2719 millennia. Both round to 49.3 millennia. This is the idealized model’s elapsed time, not a calibrated laboratory date.

b) To find the current rate of decay, we must take the derivative with respect to time, then plug in the current time relative to when the decay started. First let’s take the derivative of the function.

$M = 4 e^{- 0.1216 \cdot t}$
$\frac{d M}{d t} = - 0.1216 \cdot 4 e^{- 0.1216 \cdot t}$ (chain rule)

At the current mass M = 0.01 g, use M′ = −rM. With the displayed r ≈ 0.1216 per millennium, M′ ≈ −0.1216 · 0.01 = −0.001216 g per millennium = −1.216 mg per millennium. The negative sign means the mass is decreasing; the positive magnitude is the decay rate.

The negative simply means that the mass is decreasing.

Interactive 5.4.1 Sample Problem 5.4.1

Magnification 1×

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Model positivity and measurement limits

In the idealized exponential model, M(t) stays positive at every finite time and approaches 0 as t grows. That does not mean an arbitrarily old sample has measurable carbon-14. Real radiocarbon dating needs a remaining isotope ratio, an initial-ratio model, and calibration, and is used for once-living material only up to about 50,000 years—not the age of the non-avian dinosaurs. Our supplied initial and present amounts define a mathematics model, not a laboratory date. See the National Park Service explanation of radiocarbon dating.

Half-Life 1

On the y-axis we have the ratio of final to initial mass. Notice that when 50% of the mass has depleted, the x-value (time) is approximately equal to 5.7 millennia

Half-Life 2

Again notice that the amount of time to deplete half the previous amount has stayed the same.

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