Date post: | 07-Jul-2018 |
Category: |
Documents |
Upload: | francistsk1 |
View: | 231 times |
Download: | 0 times |
of 25
8/19/2019 Problem Set 9 Solutions.pdf
1/25
MH1100/MTH112: Calculus I.
Problem set #11.Solutions
Problem 1: (#6.1.9 to #6.1.14 from [Stewart].)
Which of the following functions are 1-1. Briefly justify.
(i) f (x) = x2 − 2x.(ii) f (x) = 10 − 3x.
(iii) g(x) = 1/x.
(iv) g(x) = |x|.
(v) h(x) = 1 + cos x.
(vi) h(x) = 1 + cos x, for 0 ≤ x ≤ π.Solution
We’ll go through these one by one.
(i) The graph of this function is a shifted parabola, so it is clearly not1-1. Another way to see this is to observe that f (0) = 0 = f (2).
(ii) The graph of this function is a non-horizontal straight line, so it willclearly pass the horizontal line test. Another way to see that f (x) is1-1 is to observe that f ′(x) = −3, so this is a decreasing function.
(iii) The graph of g(x) = 1/x is a standard hyperbola, which is clearly 1-1.Another way to see that this function is 1-1 is to deduce that if x1and x2 are two points where g(x1) = g(x2), then 1/x1 = 1/x2, whichimplies that x1 = x2.
(iv) The function g(x) = |x| is not 1-1, because g(−x) = g(x) for all x ∈ R.(v) The function h(x) = 1 + cos x is not 1-1 because for every x ∈ R,
h(x + 2π) = h(x).
(vi) This h(x) is 1-1. A transparent way of seeing this is to note thath′(x) = − sin x, which is < 0 when 0 < x < π. Thus h(x) is adecreasing function on this interval.
1
8/19/2019 Problem Set 9 Solutions.pdf
2/25
Problem 2: (Based on #6.1.19 from [Stewart].)
If h(x) = x +√
x, then what is h−1(6)? What about h
h−1(π)
?
Solution
To begin, note that both x and √ x are increasing functions, so this functionh(x) is certainly 1-1, and so the inverse function h−1(x) is logically defined.It also follows from the IVT that 6 lies in the range of h(x).
To determine h−1(6) we just have to answer the question: Which numberx̃ gives h(x̃) = 6?
Substituting in guesses, we can quickly find that h(4) = 6. A more sys-tematic approach would compute as follows. (Below, note that the domainof h is [0, ∞), so we can assume that x̃ ≥ 0.)
h(x̃) = 6
⇒ x̃ +
√ x̃ = 6
⇒√ ̃x2 + 2 ∗ 1
2√ ̃x
+ 1
4 − 25
4 = 0
⇒√
x̃ + 1
2
2=
25
4
⇒√
x̃ = 2
⇒ x̃ = 4.
The second question is what is h
h−1(π)
. Without a single calculationwe can say that the answer is π. This is because this number is the answerto the question: “What is the value in h(x) of the number whose value in
h(x) is π?”
Problem 3:
Consider f (x) = x + cos x. Explain why f (x) is 1-1. What is f −1(1)?
Solution:
The function is 1-1 because f ′(x) = 1 − sin x is always greater than zero(except at isolated points) so f (x) is an increasing function.
The value f −1
(1) is just the number x̃ which answers the question:Which number x̃ gives f (x̃) = 1? If we just think about this function (per-haps substituting in a few test values) we’ll quickly realize that f (0) = 1.Thus: f −1(1) = 0.
2
8/19/2019 Problem Set 9 Solutions.pdf
3/25
Problem 4: (#6.1.22 from [Stewart].)
In the theory of relativity, the mass of a particle with speed v is
m = f (v) = m0
1 − v2c2
.
where m0 is the “rest mass” of the particle and c is the speed of lightin a vacuum. What is the inverse function f −1(m) of f and what is itsinterpretation?
Solution:
First of all note that the domain of the function is [0, c]. (This is not surpris-ing! Nothing travels faster than light, right?) Then note that the functionis increasing. (This is easy to check formally. Or we can intuitively reason:
“As v increases from 0 to c, v2
c2 gets closer to 1, so the bottom line gets
smaller, so the whole function increases.”) So it will be 1-1 and there willbe no problems discussing an inverse function.
To actually build the inverse function: The value of f −1(m) will be thev which solves the equation:
f (v) = m
⇒ m = m0
1 − v2c2
⇒ 1 −m0
m
2=
v2
c2
⇒ v = c
1 −m0
m
2.
Thus:
f −1(m) = c
1 −
m0m
2.
The “interpretation” of this function, is that given a body with rest mass
m0, the speed it has to be going at to acquire a mass m is c
1 − m0m 2.
3
!"#$%&
8/19/2019 Problem Set 9 Solutions.pdf
4/25
Problem 5: (#6.1.27 from [Stewart].)
Find a formula for the inverse function f −1(x) to the function
f (x) = 1 − √ x1 +
√ x
.
Solution:
If we write this function
f (x) = 1
1 +√
x − 1
1√ x
+ 1
we see that f (x) is the sum of two decreasing functions, so it is decreasing.Thus f (x) is 1-1, and has a logically defined inverse. Furthermore, note thatlimx→∞ f (x) = −1. Thus we deduce that the range of f (x) is [1, −1), whichwill be the domain of the inverse function.
To determine the rule for the inverse, let x ∈ [1, −1). Then x̃ = f −1(x)satisfies:
x = 1 −
√ x̃
1 +√ ̃
x
⇒ x + x√
x̃ = 1 −√
x̃
⇒ (x + 1) √ ̃x = 1 − x⇒
√ x̃ =
1 − x1 + x
⇒ x̃ =
1 − x1 + x
2.
Thus:
f −1(x) =
1 − x1 + x
2.
4
8/19/2019 Problem Set 9 Solutions.pdf
5/25
Problem 6: (#6.1.42 from [Stewart].)
Why is
f (x) =
x3 + x2 + x + 1
1-1? Determine (f −1)′(2).
Solution
Why is f (x) a 1-1 function? First, note that x3+x2+x+1 is an increasingfunction because its derivative is 3x2 + 2x + 1 = 3(x + 1/3)2 + 2/3, which isalways > 0. Thus f (x) is the composition of two increasing functions, so itis increasing as well, and so it is 1-1.
To use the formula for the derivative of an inverse, we’ll need to knowf −1(2), so must find an x̃ such that
4 = x̃3 + x̃2 + x̃ + 1 ⇒ x̃3 + x̃2 + x̃ − 3 = 0.
A bit of trial and error leads to the solution x̃ = 1, so:
f −1(2) = 1.
Now we computed the requested derivative using the standard formulafor the derivative of an inverse:
(f −1)′(2) = 1f ′(f −1(2))
= 1
f ′(1)
= 1
12√ x3+x2+x+1
∗ (3x2 + 2x + 1)
x=1
= 1
12√ 4 ∗ (3 ∗ 1 + 2 + 1)
= 11
4 ∗6
= 2
3.
5
8/19/2019 Problem Set 9 Solutions.pdf
6/25
Problem 7: (#6.1.46 from [Stewart].)
Suppose that g(x) is the inverse function of a diff erentiable function f (x),and let G(x) = 1g(x) . If f (3) = 2 and f
′(3) = 19 , find G′(2).
Solution
We’ll compute using the standard formula for the derivative of an inverse,and other standard formulae. The computation will need g(2). Because g(x)is the inverse of f (x), g(2) is the number x̃ such that f (x̃) = 2. From thegiven datum that f (3) = 2, we deduce g (2) = 3. Now:
G′(2) = d
dx 1
g(x)
x=2
=
− 1
(g(x))2 ∗ d
dx [g(x)]
x=2
=
− 1
(g(x))2 ∗ 1
f ′(g(x))
x=2
= − 1(g(2))2
∗ 1f ′(g(2))
=
−
1
32
∗
1
f ′(3)
= −19 ∗ 1
19
= −1.
6
8/19/2019 Problem Set 9 Solutions.pdf
7/25
Problem 8: (#6.1.49 from [Stewart].)
1. Find an expression for the inverse of a function g(x) = f (x + c), wheref (x) is a one-to-one function. Interpret your result geometrically interms of the graph of a function and it’s inverse.
2. Find an expression for the inverse of h(x) = f (cx), where c̸ = 0.
Solution to (i)
To begin, note that if f (x) is 1-1, then g(x) = f (x + c) is 1-1 as well.(What would a proof look like? Assume that x1 and x2 are two numberswhere g(x1) = g(x2). This implies that f (x1 + c) = f (x2 + c). Because f (x)is 1-1 this implies that x1 + c = x2 + c, which implies that x1 = x2.)
To determine a formula for g−1(x): note that g−1(x) is the unique x̃such that g (x̃) = x. Thus:
g(x̃) = x
⇒ f (x̃ + c) = x⇒ x̃ + c = f −1(x)⇒ x̃ = f −1(x) − c.
So:g−1(x) = f −1(x) − c.
This is exactly what our geometric understanding of inverse functionspredicts: The graph of g(x) is obtained from the graph of f (x) by shifting itc units to the left. So when we reflect this arrangement over the line y = x,the graph of g(x) is reflected to the graph of g−1(x), the graph of f (x) isreflected to the graph of f −1(x), and we expect that the graph of g−1(x) isobtained from the graph of f −1(x) shifting it down c units.
Solution to (ii).
We argue in the same way:
h(x̃) = x
⇒ f (cx̃) = x⇒ cx̃ = f −1(x)⇒ x̃ = 1
cf −1(x).
Thus: h−1(x) = 1cf −1(x).
7
8/19/2019 Problem Set 9 Solutions.pdf
8/25
Problem 9: (#6.2.17 and #6.2.18 from [Stewart].)
Solution to (i).
The two points we are given determine two equations for the constants C and a:
1. 6 = C a1.
2. 24 = C a3.
If we divide the second equation by the first we deduce: a2 = 4. Thusa = 2. (We do not discuss exponential functions with negative base.)
Then substituting a = 2 into the first equation gives C = 3.Thus: f (x) = 3 ∗ 2x.
Solution to (ii).
In this case the two equations we get are:
1. 2 = C a0.
2. 2/9 = Ca2.
These can be solved to deduce: f (x) = 2 13
x, which can be rewritten
f (x) = 2 ∗ 3−x
.
8
8/19/2019 Problem Set 9 Solutions.pdf
9/25
Problem 10⋆: (#6.1.50 from [Stewart].)
Let f (x) be a 1-1, twice diff erentiable function with inverse function g(x).Show that:
g′′(x) = − f ′′(g(x))
[f ′(g(x))]3.
Solution
According to the standard formula, the derivative of g(x) is:
g′(x) = 1
f ′(g(x)).
If we diff erentiate this formula we get:
g′′(x) = − 1(f ′(g(x)))2
∗ f ′′(g(x)) ∗ g′(x)
= − 1(f ′(g(x)))2
∗ f ′′(g(x)) ∗ 1f ′(g(x))
= − 1(f ′(g(x)))3
∗ f ′′(g(x))
which is the formula we had to show.
9
8/19/2019 Problem Set 9 Solutions.pdf
10/25
Problem 11:
Use geometric transformations to sketch the graph of the function
f (x) = 2 + 5(1 − e−x).
Solution
We seek a sequence of elementary transformations which will take the stan-dard exponential function f (x) = ex into this function.
The sequence of transformations, and the corresponding geometric trans-formations, are:
!5 0 5!1
0
1
2
3
4
5
6
7
8
9
10
f (x) = ex
f (x)→f (−x)−→
!5 0 5!1
0
1
2
3
4
5
6
7
8
9
10
f (x) = e−x
f (x)→−f (x)−→
!5 0 5!10
!9
!8
!7
!6
!5
!4
!3
!2
!1
0
1
f (x) = −e−x
10
8/19/2019 Problem Set 9 Solutions.pdf
11/25
f (x)→f (x)+1−→
!5 0 5!10
!9
!8
!7
!6
!5
!4
!3
!2
!1
0
1
2
3
f (x) = 1 − e−x
f (x)→5f (x)−→
!5 0 5!20
!15
!10
!5
0
5
10
f (x) = 5(1
−e−x)
f (x)→f (x)+2−→
!5 0 5!20
!18
!16
!14
!12
!10
!8
!6
!4
!2
0
2
4
6
8
10
f (x) = 2 + 5(1 − e−x
)
11
8/19/2019 Problem Set 9 Solutions.pdf
12/25
Problem 12: (#6.2.25 from [Stewart].)
Determine the limit
limx→∞
e3x − e−3xe3x + e−3x
.
Solution
This limit can be reduced to the standard limit limx→∞ e−x = 0 by thededuction:
limx→∞
e3x − e−3xe3x + e−3x
= limx→∞
e3x
e3x ∗ 1 − e
−6x
1 + e−6x
= 1 − limx→∞ e−6x1 + limx→∞ e−6x
= 1
1
= 1.
Problem 13: (#6.2.29 from [Stewart].)
Use a version of the squeeze theorem to determine the limit
limx→∞
e−2x cos x
.
Solution
We have−1 ≤ cos x ≤ 1.
Because e−2x > 0 for all x, we can deduce from the above inequalitiesthat
−e−2x ≤ e−2x cos x ≤ e−2x.Now because
limx→∞
e−2x = limx→∞
(−e−2x) = 0we deduce form the natural version of the squeeze theorem as x → ∞ thatthe given limit limx→∞
e−2x cos x
exists, and equals 0.
12
8/19/2019 Problem Set 9 Solutions.pdf
13/25
Problem 14: (#6.2.67 from [Stewart].)
Find the absolute maximum and absolute minimum of
f (x) = xe−x2/8
on the interval [−1, 4].Solution
We are asked for the extreme values of a continuous function f (x) = xe−x2/8
on the closed interval [−1, 4]. We can obtain them from an application of the closed interval method.
The first thing we need to do is find the critical numbers of f (x). Thefunction is diff erentiable at every point, so the critical numbers are the setof numbers at which the derivative is zero: f ′(c) = 0. The derivative is:
f ′(x) = e−x2/8 − 2x
8 ∗ x ∗ e−x2/8 =
1 − x
2
4
e−x
2/8.
The only point inside the given interval [−1, 4] at which this derivative iszero is at x = 2.
To finish we need to compare the values at the critical number x = 2with the values at the boundaries of the interval, at x =
−1 and at x = 4.
The three values are:
• f (−1) = −e−1/8 ≈ −0.88• f (2) = 2e−1/2 ≈ 1.21• f (4) = 4e−2 ≈ 0.54We conclude that on the interval [−1, 4], the given function f (x) =
xe−x2/8 achieves an absolute maximum of ≈ 1.21 at x = 2, and an absolute
minimum of ≈ −0.88 at x = −1.
13
8/19/2019 Problem Set 9 Solutions.pdf
14/25
Problem 15: (#6.2.99 from [Stewart].)
Given f (x) = 3 + x + ex, calculate
f −1′
(4).
Solution
To begin, notice that this function is the sum of increasing functions, so itis increasing, so it is 1-1 and has a well-defined inverse. The calculation of the derivative of the inverse will be a standard application of the formulafor the derivative of an inverse function. The formula will need f −1(4). If we notice that f (0) = 3 + 0 + e0 = 4 we conclude that f −1(4) = 0.
Then:
d
dxf −1(x)
x=4
= 1
f ′(f −1(4))
= 1
(1 + ex)|x=0
= 1
1 + 1
= 1
2.
Problem 16: (#6.2.100 from [Stewart].)
Evaluate the limitlimx→π
esinx − 1x − π .
Solution
The trick here is to recognize that this is a limit defining a certain derivative.Set
g(x) = esinx.
Then:
limx→π
esinx − 1x − π = limx→π
esinx − esin0x − π
= g′(π
)=
cos x esinx
x=π
= (−1)esinπ= −1.
14
8/19/2019 Problem Set 9 Solutions.pdf
15/25
Problem 17: (#6.3.35 and #6.3.36 from [Stewart].)
Solve the following equations:
(a) e2x − ex − 6 = 0.(b) ln(2x + 1) = 2 − ln x.
Solution to (a)
The equation is equivalent to
(ex
−3)(ex + 2) = 0.
An x solves this equation if and only if (ex − 3) = 0 or (ex + 2) = 0. Theonly solution is x = ln3.
Solution to (b)
Because ln (2x + 1) + ln x = ln (x(2x + 1)), we can rewrite the given equa-tion:
ln (x(2x + 1)) = 2.
Because ex is 1-1, an x solves this equation if and only if
x(2x + 1) = e2
⇔ 2x2 + x − e2 = 0⇔ 2 x + 142 − 18 − e2 = 0⇔ x = −14 ± 14
√ 8e2 + 1.
Because x must be > 0 for ln x to make sense, we take the positive solutiononly:
x = −14
+ 1
4
8e2 + 1.
15
8/19/2019 Problem Set 9 Solutions.pdf
16/25
Problem 18:
Solve the following inequalities:
(a) 2 < ln x 4.
Solution to (a).
Because ex and its inverse ln x are increasing functions, a < b if and only if ea < eb. Thus:
2 < ln x 4 if and only if 2 − 3x > ln 4. Theset of solutions to this inequality is
x < 1
3(2 − ln4).
16
8/19/2019 Problem Set 9 Solutions.pdf
17/25
Problem 19: (#6.3.55 from [Stewart].)
If f (x) =√
3 − e2x, then:(i) What is the domain of f (x)?
(ii) What is f −1, stated with its domain?
Solution
An x lies in the domain of f if and only if 3 − e2x ≥ 0. So:
dom(f ) =−∞,
1
2 ln 3
.
The domain of f −1(x) is the range of f (x), which is [0,√
3).
To determine the rule for the inverse function: let x ∈ [0, √ 3) and letx̃ ∈ −∞, 12 ln 3 such that f (x̃) = x. Then:
√ 3 − e2x̃ = x
⇔ 3 − e2x̃ = x2⇔ e2x̃ = 3 − x2⇔ 2x̃ = ln(3 − x2)
⇔ x̃ = 12 ln(3
−x2).
(Note that this deduction confirms our “guess” for the range of f (x).)Thus the rule for f −1 : [0,
√ 3) → R is:
f −1(x) = 1
2 ln(3 − x2).
17
8/19/2019 Problem Set 9 Solutions.pdf
18/25
Problem 20: (#6.3.56 from [Stewart].)
If f (x) = ln(2 + ln x), then:
(i) What is the domain of f (x)?
(ii) What is f −1, stated with its domain?
Solution
The domain of ln x is (0, ∞), so the domain of ln(2 + ln x) is the set of xsuch that 2 + ln x > 0. Thus:
dom(f ) =
e−2, ∞ .The domain of the inverse function f −1(x) is the range of f , which is all
of R.
To determine the rule for the inverse function, let x ∈ R and let x̃ ∈e−2, ∞ such that f (x̃) = x. Then:
ln(2 + ln x̃) = x⇔ 2 + ln x̃ = ex⇔ x̃ = e(ex−2).
Thus the rule for the inverse function f −1
: R → R isf −1(x) = e(e
x−2).
18
8/19/2019 Problem Set 9 Solutions.pdf
19/25
Problem 21: (#6.3.61 from [Stewart].)
If f (x) = e(x3), then what is f −1(x), the corresponding inverse function?
Solution
The domain of f is all of R. The domain of the inverse function f −1(x)
is the range of f , which is (0, ∞). To determine the rule for the inversefunction, let x ∈ (0, ∞), and let x̃ ∈ R such that f (x̃) = x. Then:
x = e(x̃)3
⇔ (x̃)3 = ln x
⇔ x̃ = (ln x) 13 .
Thus the rule for the inverse function f −1 : (0, ∞) → R is:
f −1(x) = (ln x)1/3 .
19
8/19/2019 Problem Set 9 Solutions.pdf
20/25
Problem 22: (#6.3.64 from [Stewart].)
If f (x) = ex
1+2ex , then what is f −1(x), the corresponding inverse function?
Solution
To begin, note that this function does not obviously appear to be 1-1. Tounderstand it better, let’s calculate the derivative, which is
f ′(x) = ex(1 + 2ex) − ex ∗ 2ex
(1 + 2ex)2 =
ex
(1 + 2ex)2.
This is always > 0, so the function is increasing, and so is 1-1 and has a
well-defined inverse.The domain of f (x) is all of R. To understand the range, note that
limx→−∞
ex
1 + 2ex = 0
while
limx→+∞
ex
1 + 2ex = lim
x→∞1
e−x + 2 =
1
2.
Thus the range of f (x) is
0, 12
.
The domain of the inverse function is the range of f , so is 0, 12.
To determine the rule for the inverse function, let x ∈ 0, 12, and letx̃ ∈ R such that f (x̃) = x. Then:
x = ex̃
1+2ex̃
⇔ x(1 + 2ex̃) = ex̃⇔ ex̃(1 − 2x) = x⇔ x̃ = ln
x1−2x
.
Thus the inverse function f −1 :
0, 12
→ R is given by the rule
f −1(x) = ln
x1 − x
.
20
8/19/2019 Problem Set 9 Solutions.pdf
21/25
Problem 23⋆: (#6.3.67 from [Stewart].)
Consider the function
f (x) = ln
x +
x2 + 1
.
Show that f (x) is an odd function of x. Then find the corresponding inversefunction.
Solution
To begin: note that f (x) is constructed from increasing functions, so it is
itself increasing, and 1-1.The first thing we are asked to show is that f (x) is odd: f (−x) = −f (x).
Observe:
f (−x) = ln−x +
(−x)2 + 1
= ln
−x +
x2 + 1 ∗ x +
√ x2 + 1
x +√
x2 + 1
= ln−x2 + (x2 + 1)
x +√
x2 + 1
= ln
1
x + √ x2 + 1= − ln
x +
x2 + 1
= −f (x).
What is the domain of f (x)? The issue is that ln x is only defined whenx > 0. Because x +
√ x2 + 1 ≥ 1 for all x ≥ 0, we deduce that [0, ∞) lies in
the domain of f (x). Also: we just showed that f (−x) = f (x), so (−∞, 0]must lie in the domain of f (x) of f as well. Thus: the domain of f (x) is thewhole of R.
What is the range of f (x)? Note that f (0) = 0, f (x) is increasing, and
limx→∞ f (x) = ∞, which implies that [0, ∞) lies in the range of f (x). Also:f (x) is odd, as we just showed, so (−∞, 0] must lie in the range of f (x) aswell. So the range of f is all of R.
Thus the inverse function is a function with domain R and range R. To
21
8/19/2019 Problem Set 9 Solutions.pdf
22/25
determine the rule, let x and x̃
∈R be reals such that f (x̃) = x. Then:
ln
x̃ +√
x̃2 + 1
= x
⇔√
x̃2 + 1 = ex − x̃⇔ x̃2 + 1 = (e2x − 2exx̃ + x̃2)⇔ 2exx̃ = e2x − 1⇔ x̃ = 12 (ex − e−x) .
So:
f −1(x) = 1
2
ex − e−x .
22
8/19/2019 Problem Set 9 Solutions.pdf
23/25
Problem 24: (#6.3.70 from [Stewart].)
Analyse the given limits by using the equation g(x) = eln g(x) to push all thedependence on x into the index.
Solutions
Solution to (a).
limx→∞
xlnx = limx→∞
e(lnx)
lnx= lim
x→∞e(lnx)
2
= ∞.
In this analysis we have used the fact that if limx→∞ f (x) = ∞, and
limx→∞ g(x) = ∞, then limx→∞ f (g(x)) = ∞ as well. We haven’t provedthis explicitly, although it is exactly what you would expect from such limits.
Solution to (b).
limx→0+
x− lnx = limx→0+
e−(lnx)2
= 0.
In this limit, the property we have used is that if limx→−∞ f (x) = L, andlimx→0+ g(x) = −∞, then limx→0+ f (g(x)) = L. Again, this is an expectedproperty which we haven’t proved explicitly.
Solution to (c).
limx→0+
x1/x = limx→0+
e1
x∗lnx = 0.
The reasoning here is that as x → 0+, 1/x → +∞ and ln x → −∞, sotheir product tends to −∞. And because limx→−∞ ex = 0, the whole limitapproaches zero.
Solution to (d).
limx→∞
(ln2x)− lnx = limx→∞
e− lnx∗ln(ln2x) = 0.
This is a variation on the logic in Part (c).
Curious students are invited to construct proofs of the “intuitive” prop-erties we have used in the above discussions.
23
8/19/2019 Problem Set 9 Solutions.pdf
24/25
Problem 25: (#6.4.37 from [Stewart].)
Find an equation of the tangent line to the curve
y = ln (x2 − 3x + 1)at the point (3, 0).
Solution
This will just be a simple calculation. We’ll need the derivative of thefunction at the point 3.
dy
dx =
1
x2 − 3x + 1 ∗ (2x − 3).
At x = 3 this derivative is: 3.So the tangent line is given by:
y = 0 + 3(x − 3) = 3x − 9.
Problem 26: (#6.4.46 from [Stewart].)
Use logarithmic diff erentiation to diff erentiate the function
f (x) =√
xex2−x(x + 1)2/3.
Solution
The first step in logarithmic diff erentiation is to take the logarithm of bothsides:
ln f (x) = ln√
xex2−x(x + 1)2/3
.
The second step is to use the properties of the logarithm to simplify theexpression:
ln f (x) = 1
2 ln x + (x2 − x) + 2
3 ln (x + 1).
The third step is to diff erentiate both sides, using the chain rule:
f ′(x)f (x)
= 1
2x + 2x − 1 + 2
3(x + 1).
Finally, solve for the derivative you want, substituting back in f (x):
f ′(x) =
1
2x + 2x − 1 + 2
3(x + 1)
√ xex
2−x(x + 1)2
3 .
24
8/19/2019 Problem Set 9 Solutions.pdf
25/25
Problem 27: (#6.4.47 from [Stewart].)
Use logarithmic diff erentiation to diff erentiate the function
f (x) = xx.
Solution
We’ll do this one a bit more quickly:
f (x) = xx
⇒ ln f (x) = ln (xx)
⇒ ln f (x) = x ln x⇒ f ′(x)
f (x) = ln x + xx
⇒ f ′(x) = (ln x + 1) xx.
Problem 28: (#6.4.52 from [Stewart].)
Use logarithmic diff erentiation to diff erentiate the function
f (x) = (sin x)lnx.
Solution
Compute:
f (x) = (sin x)lnx
⇒ ln f (x) = ln sin xlnx⇒ ln f (x) = ln x ∗ ln (sin x)⇒ f ′(x)
f (x) = 1x ln (sin x) + ln x ∗ 1sinx ∗ cos x
⇒ f
′(x) = ln (sin x)
x + lnx cosx
sinx (sin x)ln x .
25