Math 147 Exam II Practice Problems
This review should not be used as your sole source for preparation for the exam. You
should also re-work all examples given in lecture, all homework problems, all lab assignment
problems, and all quiz problems.
1. If f (x) = x4 − 2x3 + 4x2 − 10x + 1, find the equation of the tangent line to the graph
of y = f (x) at the point (1, −6). Express your answer in slope-intercept form.
2. If f (x) = 2x3 − 3x2 − 6x + 8, find the equation of the normal line to the graph of
y = f (x) at the point (1, 1). Express your answer in slope-intercept form.
3. Differentiate f (x) = (x4 + 2x + 1)(3x2 − 5) and simplify completely.
4. Differentiate f (x) =
3x − 7
and simplify completely.
x2 + 5x − 4
5. Suppose that f (5) = 1, f 0 (5) = 6, g(5) = −3, and g 0 (5) = 2. Find the values of:
(a) (f g)0 (5)
(b) (f /g)0 (5)
6. If y = u3 + u2 + 1, where u = 2x2 − 1, find
dy
at x = 2.
dx
7. Differentiate f (x) = (x2 + 4x + 6)5 .
√
8. Differentiate f (x) = x2 − 7x. Express your answer using positive exponents.
9. Differentiate f (t) =
(2t2
3
Express your answer using positive exponents.
− 6t + 1)8
10. Differentiate f (t) = (6t2 + 5)3 (t3 − 7)4 and simplify completely.
3
x−6
11. Differentiate f (x) =
and simplify completely.
x+7
p
√
12. Differentiate f (x) = 3 1 + x. Express your answer using positive exponents.
13. Suppose that F (x) = f (g(x)), where g(3) = 6, g 0 (3) = 4, f 0 (3) = 2, and f 0 (6) = 7.
Find F 0 (3).
14. Differentiate f (x) = x csc x and simplify completely.
15. Differentiate f (x) =
sin x
and simplify completely.
1 + cos x
16. Differentiate f (x) = sin(x3 ) + cos3 x and simplify completely.
17. Differentiate f (x) = sin(cos(tan x)) and simplify completely.
1
18. Differentiate f (x) = sec2 (2x) + cot
√
1 + x2 and simplify completely.
19. Differentiate f (x) = etan x .
20. Differentiate f (x) = e−4x sin(5x) and simplify completely.
21. Differentiate f (x) =
e3x
and simplify completely.
1 + ex
2
22. Differentiate f (x) = 10x .
√
23. Differentiate f (x) = ln x. Express your answer using positive exponents.
24. Differentiate f (x) = ln(sin x) and simplify completely.
25. Differentiate f (x) = x2 ln(x3 − 4) and simplify completely.
26. Differentiate f (x) =
1 − ln x
and simplify completely.
1 + ln x
27. Differentiate f (t) = log2 (t4 − t2 + 1).
28. Differentiate f (x) = log(2x + sin x).
29. Differentiate f (x) = sin−1 (x2 − 1) and simplify completely.
√
30. Differentiate f (x) = x arctan x and simplify completely.
√
31. Differentiate f (t) = cos−1 2t − 1 and simplify completely.
df −1
32. If f (x) = 2x + cos x, find
(1) = (f −1 )0 (1).
dx
√
df −1
3
2
33. If f (x) = x + x + x + 1, find
(2) = (f −1 )0 (2).
dx
34. Consider the curve defined implicitly by
x3 + y 3 = 6xy
dy
.
dx
(b) Find the equation of the tangent line to the curve at the point (3, 3). Express
your answer in slope-intercept form.
(a) Find
35. Consider the curve defined implicitly by
x cos y + y cos x = 1
dy
.
dx
(b) Find the equation of the tangent line to the curve at the point (0, 1). Express
your answer in slope-intercept form.
(a) Find
2
36. Consider the curve defined implicitly by
cos(x − y) = xex
(a) Find y 0 = dy/dx.
(b) Find the equation of the tangent line to the curve at the point (0, π/2). Express
your answer in slope-intercept form.
37. Consider the curve defined implicitly by
y = ln(x2 + y 2 )
(a) Find y 0 = dy/dx.
(b) Find the equation of the tangent line to the curve at the point (1, 0). Express
your answer in slope-intercept form.
38. Use logarithmic differentiation to find the derivative of
√
(x3 + 1)4 x5 + 2
y=
(x + 1)3 (x2 + 3)2
39. Use logarithmic differentiation to find the derivative of f (x) = (sin x)cos x .
40. Find all higher derivatives of f (x) = x6 − 2x5 + 3x4 − 4x3 + 5x2 − 6x + 7.
41. If f (x) =
1
, find f (2016) (x).
x3
42. If f (x) = sin(2x), find f (50) (x).
43. Find the thousandth derivative of f (x) = xe−x .
44. If f (x) = ln(x − 1), find f (99) (x).
45. Find y 00 if x4 + y 4 = 16.
46. The position of a particle is given by the equation
s(t) = t3 − 6t2 + 9t
where t is measured in seconds and s in feet.
(a) Find the velocity and acceleration at time t.
(b) What are the velocity and acceleration after 2 s? Include the appropriate units.
(c) Find the average velocity from t = 1 to t = 5 s. Include the appropriate unit for
velocity.
(d) When is the particle at rest? What is the acceleration at these times? Include
the appropriate unit for acceleration.
3
47. The position of a particle is given by the equation
√
s(t) = t2 + t + 4
where s is measured in meters and t in seconds.
(a) Find the average velocity of the particle from t = 0 to t = 3 s. Include the
appropriate unit for velocity.
(b) Find the instantaneous velocity of the particle after 3 s. Include the appropriate
unit for velocity.
(c) State the name of a theorem which guarantees that there is some time T ∈ (0, 3)
such that the instantaneous velocity of the particle at time T seconds is equal
to the average velocity found in part (a)? Find the time T guaranteed by this
theorem.
48. Air is being pumped into a spherical balloon so that its volume increases at a rate
of 100 cubic centimeters per second. How fast is the radius of the balloon increasing
when the diameter is 50 centimeters?
49. A ladder 10 ft long rests against a vertical wall. The bottom of the ladder slides away
from the wall at a rate of 2 ft/s.
(a) How fast is the top of the ladder sliding down the wall when the bottom of the
ladder is 4 ft from the wall?
(b) How fast is the angle between the top of the ladder and the wall changing when
the angle is π/4 radians?
50. A water tank has the shape of an inverted circular cone with base radius 2 m and
height 4 m. If water is being pumped into the tank at a rate of 2 m3 /min, find the
rate at which the water level is rising when the water is 3 m deep.
51. Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are
headed for the intersection of the two roads. At what rate are the cars approaching
each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
√
52. Use a linear approximation to find an approximate value for 3 64.1.
53. Find the absolute extrema of f (x) = x3 − 3x2 + 1 on [−1, 3].
54. Find the value of c which satisfies the Mean Value Theorem for f (x) =
x
on the
x+5
interval [1, 10].
55. Suppose that f (x) is continuous on the interval [2, 5] and differentiable on the interval
(2, 5). Show that if 1 ≤ f 0 (x) ≤ 4 for all x ∈ [2, 5], then 3 ≤ f (5) − f (2) ≤ 12.
4
Solutions
Solutions may contain errors or typos. If you find an error or typo, please notify me at
[email protected].
1. If f (x) = x4 − 2x3 + 4x2 − 10x + 1, find the equation of the tangent line to the graph
of y = f (x) at the point (1, −6). Express your answer in slope-intercept form.
The slope is m = f 0 (1), which we calculate as follows:
f 0 (x) = 4x3 − 6x2 + 8x − 10
f 0 (1) = 4 − 6 + 8 − 10 = −4
Therefore, the equation of the tangent line is
y + 6 = −4(x − 1)
y = −4x − 2
2. If f (x) = 2x3 − 3x2 − 6x + 8, find the equation of the normal line to the graph of
y = f (x) at the point (1, 1). Express your answer in slope-intercept form.
The slope of the tangent line is m1 = f 0 (1) which we calculate as follows:
f 0 (x) = 6x2 − 6x − 6
f 0 (1) = −6
Thus, the slope of the normal line is m2 = −1/m1 = 1/6. Therefore, the equation of
the tangent line is
1
(x − 1)
6
1
1
y−1 =
x−
6
6
1
5
y =
x+
6
6
y−1 =
5
3. Differentiate f (x) = (x4 + 2x + 1)(3x2 − 5) and simplify completely.
By the Product Rule, we have
d
d
(3x2 − 5) + (3x2 − 5) (x4 + 2x + 1)
dx
dx
4
2
3
= (x + 2x + 1)(6x) + (3x − 5)(4x + 2)
= 6x5 + 12x2 + 6x + 12x5 + 6x2 − 20x3 − 10
= 18x5 − 20x3 + 18x2 + 6x − 10
f 0 (x) = (x4 + 2x + 1)
Notice that we could first multiply the factors:
f (x) = (x4 + 2x + 1)(3x2 − 5)
= 3x6 + 6x3 + 3x2 − 5x4 − 10x − 5
= 3x6 − 5x4 + 6x3 + 3x2 − 10x − 5
and then differentiate to obtain
f 0 (x) = 18x5 − 20x3 + 18x2 + 6x − 10
4. Differentiate f (x) =
x2
3x − 7
and simplify completely.
+ 5x − 4
By the Quotient Rule, we have
(x2 + 5x − 4)D(3x − 7) − (3x − 7)D(x2 + 5x − 4)
(x2 + 5x − 4)2
(x2 + 5x − 4)(3) − (3x − 7)(2x + 5)
=
(x2 + 5x − 4)2
3x2 + 15x − 12 − (6x2 + x − 35)
=
(x2 + 5x − 4)2
−3x2 + 14x + 23
=
(x2 + 5x − 4)2
f 0 (x) =
6
5. Suppose that f (5) = 1, f 0 (5) = 6, g(5) = −3, and g 0 (5) = 2. Find the values of:
(a) (f g)0 (5)
By the Product Rule, we have
(f g)0 (5) = f 0 (5)g(5) + f (5)g 0 (5)
= 6(−3) + 1(2) = −16
(b) (f /g)0 (5)
By the Quotient Rule, we have
f 0 (5)g(5) − f (5)g 0 (5)
[g(5)]2
20
6(−3) − 1(2)
= −
=
2
(−3)
9
(f /g)0 (5) =
6. If y = u3 + u2 + 1, where u = 2x2 − 1, find
dy
at x = 2.
dx
By the Chain Rule, we have
dy
dy du
=
= (3u2 + 2u)(4x)
dx
du dx
At x = 2, u = 2(2)2 − 1 = 7, and so
dy
= (3 · 72 + 2 · 7)(4 · 2) = 161 · 8 = 1288
dx
7. Differentiate f (x) = (x2 + 4x + 6)5 .
By the Chain Rule, we have
d 2
(x + 4x + 6)
dx
= 5(x2 + 4x + 6)4 (2x + 4)
= (10x + 20)(x2 + 4x + 6)4
f 0 (x) = 5(x2 + 4x + 6)4
7
√
8. Differentiate f (x) =
x2 − 7x. Express your answer using positive exponents.
We first rewrite f as
f (x) = (x2 − 7x)1/2
By the Chain Rule, we have
d
1 2
(x − 7x)−1/2 (x2 − 7x)
2
dx
1
= √
(2x − 7)
2 x2 − 7x
2x − 7
= √
2 x2 − 7x
f 0 (x) =
9. Differentiate f (t) =
(2t2
3
Express your answer using positive exponents.
− 6t + 1)8
We first rewrite f as
f (t) = 3(2t2 − 6t + 1)−8
By the Chain Rule, we have
d
(2t2 − 6t + 1)
dt
= −24(2t2 − 6t + 1)−9 (4t − 6)
24(4t − 6)
= − 2
(2t − 6t + 1)9
f 0 (t) = −24(2t2 − 6t + 1)−9
10. Differentiate f (t) = (6t2 + 5)3 (t3 − 7)4 and simplify completely.
By the Product and Chain Rules, we have
d
d
[(6t2 + 5)3 ](t3 − 7)4 + (6t2 + 5)3 [(t3 − 7)4 ]
dt
dt
= 3(6t2 + 5)2 (12t)(t3 − 7)4 + (6t2 + 5)3 · 4(t3 − 7)3 (3t2 )
= 36t(6t2 + 5)2 (t3 − 7)4 + 12t2 (6t2 + 5)3 (t3 − 7)3
f 0 (t) =
By using common factors, we can simplify the answer as
f 0 (t) = 12t(6t2 + 5)2 (t3 − 7)3 [3(t3 − 7) + t(6t2 + 5)]
= 12t(6t2 + 5)2 (t3 − 7)3 (9t3 + 5t − 21)
8
11. Differentiate f (x) =
x−6
x+7
3
and simplify completely.
By the Chain and Quotient Rules, we have
2
x−6
d x−6
0
f (x) = 3
x+7
dx x + 7
2
x − 6 (1)(x + 7) − (x − 6)(1)
= 3
x+7
(x + 7)2
2
13
3(x − 6)
=
2
(x + 7) (x + 7)2
39(x − 6)2
=
(x + 7)4
12. Differentiate f (x) =
p
3
1+
√
x. Express your answer using positive exponents.
We first rewrite f as
f (x) = (1 +
√
x)1/3
By the Chain Rule, we have
√
√
d
1
(1 + x)−2/3 (1 + x)
3
dx 1
1
√ 2/3
√
=
3(1 + x)
2 x
1
√
= √
6 x(1 + x)2/3
f 0 (x) =
Notice that the Chain Rule has been used twice.
13. Suppose that F (x) = f (g(x)), where g(3) = 6, g 0 (3) = 4, f 0 (3) = 2, and f 0 (6) = 7.
Find F 0 (3).
By the Chain Rule, we have
F 0 (3) = f 0 (g(3))g 0 (3)
= f 0 (6)(4)
= 7(4) = 28
9
14. Differentiate f (x) = x csc x and simplify completely.
By the Product Rule, we have
f 0 (x) = (1) csc x + x(− csc x cot x)
= csc x − x csc x cot x
15. Differentiate f (x) =
sin x
and simplify completely.
1 + cos x
By the Quotient Rule, we have
cos x(1 + cos x) − sin x(− sin x)
(1 + cos x)2
cos x + cos2 x + sin2 x
=
(1 + cos x)2
cos x + 1
=
(1 + cos x)2
1
=
1 + cos x
f 0 (x) =
In simplifying the answer, we have used the identity sin2 x + cos2 x = 1.
16. Differentiate f (x) = sin(x3 ) + cos3 x and simplify completely.
By the Chain Rule, we have
f 0 (x) = cos(x3 )(3x2 ) + 3 cos2 x(− sin x)
= 3x2 cos(x3 ) − 3 cos2 x sin x
17. Differentiate f (x) = sin(cos(tan x)) and simplify completely.
By the Chain Rule, we have
d
[cos(tan x)]
dx
d
= cos(cos(tan x))[− sin(tan x)] (tan x)
dx
= − cos(cos(tan x)) sin(tan x) sec2 x
f 0 (x) = cos(cos(tan x))
Notice that the Chain Rule has been used twice.
10
18. Differentiate f (x) = sec2 (2x) + cot
√
1 + x2 and simplify completely.
By the Chain Rule, we have
√
d
d √
[sec(2x)] − csc2 1 + x2 ( 1 + x2 )
dx
dx
√
1
d
2
2
= 2 sec(2x)[2 sec(2x) tan(2x)] − csc 1 + x2 √
(1 + x )
2 1 + x2 dx
√
1
2
2
= 4 sec (2x) tan(2x) − csc 1 + x2 √
(2x)
2 1 + x2
√
x
2
2
2
= 4 sec (2x) tan(2x) − csc 1 + x √
1 + x2
f 0 (x) = 2 sec(2x)
19. Differentiate f (x) = etan x .
By the Chain Rule, we have
d
(tan x)
dx
= etan x sec2 x
f 0 (x) = etan x
20. Differentiate f (x) = e−4x sin 5x and simplify completely.
By the Product and Chain Rules, we have
d −4x
d
(e ) sin 5x + e−4x (sin 5x)
dx
dx
= −4e−4x sin 5x + e−4x (5 cos 5x)
= e−4x (−4 sin 5x + 5 cos 5x)
f 0 (x) =
21. Differentiate f (x) =
e3x
and simplify completely.
1 + ex
By the Quotient and Chain Rules, we have
3e3x (1 + ex ) − e3x (ex )
(1 + ex )2
3x
3e + 3e4x − e4x
=
(1 + ex )2
3e3x + 2e4x
=
(1 + ex )2
f 0 (x) =
11
2
22. Differentiate f (x) = 10x .
By the Chain Rule, we have
d 2
(x )
dx
2
= (ln 10)10x (2x)
f 0 (x) = (ln 10)10x
23. Differentiate f (x) =
√
2
ln x. Express your answer using positive exponents.
By the Chain Rule, we have
d
1
√
(ln x)
2 ln x dx
1 1
= √
2 ln x x
1
√
=
2x ln x
f 0 (x) =
24. Differentiate f (x) = ln(sin x) and simplify completely.
By the Chain Rule, we have
f 0 (x) =
cos x
= cot x
sin x
25. Differentiate f (x) = x2 ln(x3 − 4) and simplify completely.
By the Product and Chain Rules, we have
d 2
d
(x ) ln(x3 − 4) + x2 [ln(x3 − 4)]
dx
dx
2
3x
= 2x ln(x3 − 4) + x2 3
x −4
3x4
= 2x ln(x3 − 4) + 3
x −4
f 0 (x) =
12
26. Differentiate f (x) =
1 − ln x
and simplify completely.
1 + ln x
By the Quotient Rule, we have
− x1 (1 + ln x) − (1 − ln x) x1
f (x) =
(1 + ln x)2
− x2
=
(1 + ln x)2
2
= −
x(1 + ln x)2
0
27. Differentiate f (t) = log2 (t4 − t2 + 1).
By the Chain Rule, we have
f 0 (t) =
4t3 − 2t
(t4 − t2 + 1) ln 2
28. Differentiate f (x) = log(2x + sin x).
By the Chain Rule, we have
f 0 (x) =
2 + cos x
(2x + sin x) ln 10
29. Differentiate f (x) = sin−1 (x2 − 1).
By the Chain Rule, we have
1
d 2
f 0 (x) = p
(x − 1)
1 − (x2 − 1)2 dx
1
= p
2x
4
1 − (x − 2x2 + 1)
2x
= √
2x2 − x4
13
√
30. Differentiate f (x) = x arctan x and simplify completely.
By the Product and Chain Rules, we have
√
√
d
d
(x) arctan x + x (arctan x)
dx
dx
√
1
d√
= (1) arctan x + x √ 2
x
( x) + 1 dx
√
x
1
√
= arctan x +
x+12 x
√
√
x
= arctan x +
2(x + 1)
f 0 (x) =
31. Differentiate f (t) = cos−1
√
2t − 1 and simplify completely.
By the Chain Rule, we have
1
d√
f 0 (t) = − q
2t − 1
√
dt
2
1 − ( 2t − 1)
1
2
= −p
· √
1 − (2t − 1) 2 2t − 1
1
√
= −√
2 − 2t 2t − 1
32. If f (x) = 2x + cos x, find
df −1
(1) = (f −1 )0 (1).
dx
To use the theorem for the derivative of an inverse function, we find f 0 (x) = 2 − sin x.
Additionally, we need to know f −1 (1) and we can find it by inspection. Since f (0) = 1,
f −1 (1) = 0. Therefore,
(f −1 )0 (1) =
1
f 0 [f −1 (1)]
14
=
1
1
1
=
=
f 0 (0)
2 − sin 0
2
33. If f (x) =
√
x3 + x2 + x + 1, find
df −1
(2) = (f −1 )0 (2).
dx
To use the theorem for the derivative of an inverse function, we find
3x2 + 2x + 1
f 0 (x) = √
2 x3 + x2 + x + 1
Additionally, we need to know f −1 (2) and we can find it by inspection. Since f (1) = 2,
f −1 (2) = 1. Therefore,
√
1
2 1+1+1+1
4
2
1
−1 0
= 0
=
= =
(f ) (2) = 0 −1
f [f (2)]
f (1)
3+2+1
6
3
34. Consider the curve defined implicitly by
x3 + y 3 = 6xy
(a) Find
dy
.
dx
Differentiating both sides with respect to x, regarding y as a function of x, and
using the Chain Rule on the y 3 term and Product Rule on the 6xy term, we have
3x2 + 3y 2 y 0 = 6y + 6xy 0
x2 + y 2 y 0 = 2y + 2xy 0
Solving for y 0 , we obtain
(y 2 − 2x)y 0 = 2y − x2
2y − x2
y0 = 2
y − 2x
(b) Find the equation of the tangent line to the curve at the point (3, 3). Express
your answer in slope-intercept form.
When x = y = 3, the slope of the tangent line is
y0 =
2(3) − 32
= −1
32 − 2(3)
So the equation of the tangent line is
y − 3 = −1(x − 3)
y = −x + 6
15
35. Consider the curve defined implicitly by
x cos y + y cos x = 1
(a) Find
dy
.
dx
Differentiating both sides with respect to x, regarding y as a function of x, and
using the Product and Chain Rules, we have
(1) cos y − x sin y(y 0 ) + y 0 cos x − y sin x = 0
cos y − xy 0 sin y + y 0 cos x − y sin x = 0
Solving for y 0 , we obtain
−xy 0 sin y + y 0 cos x = − cos y + y sin x
y 0 (−x sin y + cos x) = − cos y + y sin x
− cos y + y sin x
y0 =
−x sin y + cos x
(b) Find the equation of the tangent line to the curve at the point (0, 1). Express
your answer in slope-intercept form.
When x = 0 and y = 1, the slope of the tangent line is
y0 =
− cos 1 + 1 sin 0
= − cos 1
−0 sin 1 + cos 0
So the equation of the line is
y = 1 − (cos 1)x
16
36. Consider the curve defined implicitly by
cos(x − y) = xex
(a) Find
dy
.
dx
Differentiating both sides with respect to x, regarding y as a function of x, and
using the Chain Rule on the cos(x − y) term and Product Rule on the xex term,
we have
− sin(x − y)(1 − y 0 ) = (1)ex + x(ex )
− sin(x − y) + y 0 sin(x − y) = ex (1 + x)
Solving for y 0 , we obtain
y 0 sin(x − y) = sin(x − y) + ex (1 + x)
ex (1 + x)
y0 = 1 +
sin(x − y)
(b) Find the equation of the tangent line to the curve at the point (0, π/2). Express
your answer in slope-intercept form.
When x = 0 and y = π/2, the slope of the tangent line is
y0 = 1 +
e0 (1 + 0)
1
=1+
=0
sin(0 − π/2)
−1
So the tangent line is the horizontal line y = π/2.
17
37. Consider the curve defined implicitly by
y = ln(x2 + y 2 )
(a) Find
dy
.
dx
Differentiating both sides with respect to x, regarding y as a function of x, and
using the Chain Rule on the ln(x2 + y 2 ) term, we have
y0 =
2x + 2yy 0
x2 + y 2
Solving for y 0 , we obtain
y 0 (x2 + y 2 ) = 2x + 2yy 0
y 0 (x2 + y 2 ) − 2yy 0 = 2x
y 0 (x2 + y 2 − 2y) = 2x
2x
y0 = 2
x + y 2 − 2y
(b) Find the equation of the tangent line to the curve at the point (1, 0). Express
your answer in slope-intercept form.
When x = 1 and y = 0, the slope of the tangent line is
y0 =
2(1)
=2
1 + 0 − 2(0)
So the equation of the line is
y − 0 = 2(x − 1)
y = 2x − 2
18
38. Use logarithmic differentiation to find the derivative of
√
(x3 + 1)4 x5 + 2
y=
(x + 1)3 (x2 + 3)2
We take logarithms of both sides of the equation:
ln y = 4 ln(x3 + 1) +
1
ln(x5 + 2) − 3 ln(x + 1) − 2 ln(x2 + 3)
2
Differentiating implicitly with respect to x gives
y0
12x2
5x4
3
4x
= 3
+ 5
−
− 2
y
x + 1 2x + 4 x + 1 x + 3
Solving for y 0 , we obtain
12x2
5x4
3
4x
0
y = y
+
−
−
x3 + 1 2x5 + 4 x + 1 x2 + 3
√
(x3 + 1)4 x5 + 2 12x2
5x4
3
4x
=
+
−
−
(x + 1)3 (x2 + 3)2 x3 + 1 2x5 + 4 x + 1 x2 + 3
39. Use logarithmic differentiation to find the derivative of f (x) = (sin x)cos x .
Let y = (sin x)cos x . We take logarithms of both sides of the equation:
ln y = ln[(sin x)cos x ] = (cos x) ln(sin x)
Differentiating implicitly with respect to x gives
y0
cos x
= (− sin x) ln(sin x) + (cos x)
y
sin x
0
y
= −(sin x) ln(sin x) + cos x cot x
y
Solving for y 0 , we obtain
y 0 = y [−(sin x) ln(sin x) + cos x cot x]
= (sin x)cos x [−(sin x) ln(sin x) + cos x cot x]
19
40. Find all higher derivatives of f (x) = x6 − 2x5 + 3x4 − 4x3 + 5x2 − 6x + 7.
Differentiating repeatedly, we have
f 0 (x)
f 00 (x)
f 000 (x)
f (4) (x)
f (5) (x)
f (6) (x)
f (7) (x)
=
=
=
=
=
=
=
6x5 − 10x4 + 12x3 − 12x2 + 10x − 6
30x4 − 40x3 + 36x2 − 24x + 10
120x3 − 120x2 + 72x − 24
360x2 − 240x + 72
720x − 240
720
0
In fact, f (n) (x) = 0 for all n ≥ 7.
41. If f (x) =
1
, find f (2016) (x).
x3
Differentiating repeatedly, we have
f (x) = x−3 =
1
x3
f 0 (x) = −3x−4 = −
3
x4
f 00 (x) = (−4)(−3)x−5 =
12
x5
f 000 (x) = −5 · 4 · 3 · x−6
f (4) (x) = 6 · 5 · 4 · 3 · x−7
f (5) (x) = −7 · 6 · 5 · 4 · 3 · x−8 = −
7!
2x8
..
.
(n)
f (x) = (−1)n (n + 2)(n + 1)(n) · · · 4 · 3 · x−(n+3)
(−1)n (n + 2)!
f (n) (x) =
2xn+3
Therefore, the 2016th derivative is
f (2016) (x) =
20
2018!
2x2019
42. If f (x) = sin(2x), find f (50) (x).
The first few derivatives of f (x) = sin(2x) are
f 0 (x)
f 00 (x)
f 000 (x)
f (4) (x)
f (5) (x)
=
=
=
=
=
2 cos(2x)
−4 sin(2x)
−8 cos(2x)
16 sin(2x)
32 cos(2x)
We see that the successive derivatives occur in a cycle of length 4, the coefficients are
powers of 2 and, in particular,
f (n) (x) = 2n sin(2x)
whenever n is a multiple of 4. Therefore,
f (48) (x) = 248 sin(2x)
and, differentiating two more times, we have
f (50) (x) = −250 sin(2x)
43. Find the thousandth derivative of f (x) = xe−x .
The first few derivatives of f (x) = xe−x are
f 0 (x)
f 00 (x)
f 000 (x)
f (4) (x)
=
=
=
=
..
.
(n)
f (x) =
e−x − xe−x = e−x (1 − x)
−e−x (1 − x) − e−x = e−x (x − 2)
−e−x (x − 2) + e−x = e−x (3 − x)
−e−x (3 − x) − e−x = e−x (x − 4)
(−1)n e−x (x − n)
Therefore, the 1000th derivative is
f (1000) (x) = e−x (x − 1000)
21
44. If f (x) = ln(x − 1), find f (99) (x).
The first few derivatives of f (x) = ln(x − 1) are
f 0 (x) =
1
= (x − 1)−1
x−1
1
(x − 1)2
2
f 000 (x) = 2(x − 1)−3 =
(x − 1)3
3!
f (4) (x) = −3 · 2(x − 1)−4 = −
(x − 1)4
4!
f (5) (x) = 4 · 3 · 2(x − 1)−5 =
(x − 1)5
..
.
f 00 (x) = −(x − 1)−2 = −
f (n) (x) = (−1)n+1 (n − 1)!(x − 1)−n =
Therefore, the 99th derivative is
f (99) (x) =
22
98!
(x − 1)99
(−1)n+1 (n − 1)!
(x − 1)n
45. Find y 00 if x4 + y 4 = 16.
Differentiating the equation implicitly with respect to x, we obtain
4x3 + 4y 3 y 0 = 0
Solving for y 0 gives
4y 3 y 0 = −4x3
x3
y0 = − 3
y
Using the Quotient Rule and remembering that y is a function of x, we have
3x2 (y 3 ) − x3 (3y 2 y 0 )
y6
3x2 y 3 − 3x3 y 2 y 0
= −
y6
y 00 = −
Substituting the expression we obtained for y 0 into this expression, we obtain
3
−x
2 3
3 2
3x y − 3x y
y3
y 00 = −
y6
2 4
3x y + 3x6
= −
y7
2 4
3x (y + x4 )
= −
y7
But the values of x and y must satisfy the original equation x4 + y 4 = 16. So the
answer simplifies to
3x2 (16)
48x2
y 00 = −
=
−
y7
y7
23
46. The position of a particle is given by the equation
s(t) = t3 − 6t2 + 9t
where t is measured in seconds and s in feet.
(a) Find the velocity and acceleration at time t.
The velocity function is the derivative of the position function:
v(t) = s0 (t) = 3t2 − 12t + 9
The acceleration function is the derivative of the velocity function:
a(t) = v 0 (t) = s00 (t) = 6t − 12
(b) What are the velocity and acceleration after 2 s? Include the appropriate units.
The velocity after 2 s means the instantaneous velocity when t = 2. That is,
v(2) = 3(2)2 − 12(2) + 9 = −3 ft/s
Similarly, the acceleration after 2 s is
a(2) = 6(2) − 12 = 0 ft/s2
(c) Find the average velocity from t = 1 to t = 5 s. Include the appropriate unit for
velocity.
The average velocity from t = 1 to t = 5 is
Change in position
s(5) − s(1)
20 − 4
=
=
= 4 f/s
Change in time
5−1
5−1
(d) When is the particle at rest? What is the acceleration at these times? Include
the appropriate unit for acceleration.
The particle is at rest when v(t) = 0. That is,
3t2 − 12t + 9 = 3(t2 − 4t + 3) = 3(t − 1)(t − 3) = 0
and this is true when t = 1 or t = 3. Thus, the particle is at rest after 1 s and
after 3 s. The acceleration at these times is
a(1) = 6(1) − 12 = −6 ft/s2
a(3) = 6(3) − 12 = 6 ft/s2
24
47. The position of a particle is given by the equation
√
s(t) = t2 + t + 4
where s is measured in meters and t in seconds.
(a) Find the average velocity of the particle from t = 0 to t = 3 s. Include the appropriate unit for velocity.
The average velocity from t = 0 to t = 3 is
Change in position
s(3) − s(0)
4−2
2
=
=
= m/s
Change in time
3−0
3−0
3
(b) Find the instantaneous velocity of the particle after 3 s. Include the appropriate
unit for velocity.
The velocity function is the derivative of the position function. By the Chain
Rule, we have
2t + 1
v(t) = s0 (t) = √
2 t2 + t + 4
After t = 3 seconds, the instantaneous velocity is
2(3) + 1
7
v(3) = p
= m/s
8
2 (3)2 + 3 + 4
25
(c) State the name of a theorem which guarantees that there is some time T ∈ (0, 3)
such that the instantaneous velocity of the particle at time T seconds is equal
to the average velocity found in part (a)? Find the time T guaranteed by this
theorem.
Since s(t) is continuous and differentiable on the interval [0, 3], the Mean Value
Theorem guarantees that there is some time T ∈ (0, 3) such that the instantaneous
velocity of the particle at time T seconds is equal to the average velocity from
t = 0 to t = 3 s.
To calculate this time T , let
2
v(T ) =
3
Therefore, we have
2T + 1
√
2 T2 + T + 4
3(2T + 1)
6T + 3
(6T + 3)2
36T 2 + 36T + 9
20T 2 + 20T − 55
=
=
=
=
=
=
2
3
√
4 T2 + T + 4
√
4 T2 + T + 4
16(T 2 + T + 4)
16T 2 + 16T + 64
0
Using the Quadratic Formula, we obtain
p
√
−20 ± (20)2 − 4(20)(−55)
−1 ± 2 3
=
T =
2(20)
2
Since the value of T must lie in the interval (0, 3), we reject the negative value of
T and obtain
1 √
T = − + 3 ≈ 1.2 s
2
26
48. Air is being pumped into a spherical balloon so that its volume increases at a rate
of 100 cubic centimeters per second. How fast is the radius of the balloon increasing
when the diameter is 50 centimeters?
The volume V of a sphere with radius r is
4
V = πr3
3
We differentiate both sides of this equation with respect to t. Using the Chain Rule,
we have
dV
dr
dV dr
=
= 4πr2
dt
dr dt
dt
Now we solve for the unknown quantity:
1 dV
dr
=
dt
4πr2 dt
Substituting r = 25 and dV /dt = 100 in this equation, we obtain
1
1
dr
=
=
2
dt
4π(25)
25π
The radius of the balloon is increasing at the rate of 1/(25π) cm/s.
27
49. A ladder 10 ft long rests against a vertical wall. The bottom of the ladder slides away
from the wall at a rate of 2 ft/s.
(a) How fast is the top of the ladder sliding down the wall when the bottom of the
ladder is 4 ft from the wall?
We first draw a diagram and label it as in the figure below.
θ
10
y
x
By the Pythagorean Theorem,
x2 + y 2 = 100
Differentiating each side with respect to t using the Chain Rule, we have
2x
dx
dy
+ 2y
=0
dt
dt
and solving this equation for the desired rate, we obtain
dy
x dx
=−
dt
y dt
When x = 4, the Pythagorean Theorem gives y =
values and dx/dt = 2, we have
√
84 and so, substituting these
dy
4
4
= − √ (2) = − √
dt
84
21
√
The top of the ladder slides down the wall at the rate of 4/ 21 ft/s.
28
(b) How fast is the angle between the top of the ladder and the wall changing when
the angle is π/4 radians?
Using the diagram, we have
x
10
Differentiating each side with respect to t using the Chain Rule, we have
sin θ =
cos θ
dθ
1 dx
=
dt
10 dt
and solving this equation for the desired rate, we obtain
dθ
sec θ dx
=
dt
10 dt
Substituting θ = π/4 and dx/dt = 2, we have
√
sec(π/4)
2
dθ
=
(2) =
dt
10
5
The
√ angle between the top of the ladder and the wall is increasing at the rate of
2/5 rad/s.
29
50. A water tank has the shape of an inverted circular cone with base radius 2 m and
height 4 m. If water is being pumped into the tank at a rate of 2 m3 /min, find the
rate at which the water level is rising when the water is 3 m deep.
We first sketch the cone and label it as in the figure below.
2
r
4
h
The volume V of the cone of water with height h and radius r is
V =
π 2
r h
3
but it is very useful to express V as a function of h alone. In order to eliminate r we
use the similar triangles to write
r
2
=
h
4
r=
h
2
and the expression for V becomes
π
V =
3
2
h
π
h = h3
2
12
Differentiating each side with respect to t using the Chain Rule, we have
dV
π dh
= h2
dt
4 dt
and solving this equation for the desired rate, we obtain
dh
4 dV
=
dt
πh2 dt
Substituting h = 3 and dV /dt = 2, we have
dh
4
8
=
(2) =
2
dt
π(3)
9π
The water level is rising at the rate of 8/(9π) m/min.
30
51. Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are
headed for the intersection of the two roads. At what rate are the cars approaching
each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
We first draw a diagram and label it as in the figure below.
x
y
A
z
B
By the Pythagorean Theorem,
z 2 = x2 + y 2
Differentiating each side with respect to t using the Chain Rule, we have
2z
dz
dx
dy
= 2x + 2y
dt
dt dt
1
dy
dz
dx
=
x +y
dt
z
dt
dt
When x = 0.3 and y = 0.4, the Pythagorean Theorem gives z = 0.5, so
dz
1
=
[0.3(−50) + 0.4(−60)] = −78
dt
0.5
The cars are approaching each other at a rate of 78 mi/h.
31
52. Use a linear approximation to find an approximate value for
Let f (x) =
√
3
64.1.
√
3
x = x1/3 and a = 64. It follows that
f 0 (x) =
1
3x2/3
Evaluating at a = 64, we obtain f (64) = 4 and
f 0 (64) =
1
1
1
=
=
2/3
2
3(64)
3(4)
48
Therefore, the linearization of f (x) at a = 64 is
L(x) = f (64) + f 0 (64)(x − 64)
1
= 4 + (x − 64)
48
√
Thus, a linear approximation of 3 64.1 is
√
3
64.1 ≈ L(64.1)
1
= 4 + (64.1 − 64)
48 1
1
= 4+
48 10
1
= 4+
480
1921
=
480
≈ 4.0021
√
Using a calculator, the actual value of 3 64.1 is 4.00208 . . .. Thus, the approximation
is accurate to three decimal places
32
53. Find the absolute extrema of f (x) = x3 − 3x2 + 1 on [−1, 3].
To find the critical values of f , we set
f 0 (x) = 3x2 − 6x = 3x(x − 2) = 0.
Thus, the critical values are x = 0 and x = 2 which both lie in the interval [−1, 3].
Evaluating the function at the critical values and the endpoints of the interval, we
obtain
f (−1)
f (0)
f (2)
f (3)
=
=
=
=
−3
1
−3
1
Therefore, f has an absolute maximum value of 1 and and absolute minimum value of
−3. The graph of y = f (x) is given below.
33
54. Find the value of c which satisfies the Mean Value Theorem for f (x) =
x
on the
x+5
interval [1, 10].
First, note that f is continuous on the interval [1, 10] and differentiable on (1, 10). By
the Mean Value Theorem, there exists a number c ∈ (1, 10) such that
f 0 (c) =
By the Quotient Rule,
f (10) − f (1)
10 − 1
x+5−x
5
=
2
(x + 5)
(x + 5)2
f 0 (x) =
Moreover,
2
−1
1
f (10) − f (1)
= 3 6 =
10 − 1
9
18
Therefore, we want to calculate the value of c ∈ (2, 5) which satisfies
5
1
=
2
(c + 5)
18
Solving for c, we obtain
(c + 5)2
c+5
c+5
c
=
=
=
=
90
√
± 90
√
±3 10
√
±3 10 − 5
√
Since we seek the value of c which lies in the interval (1, 10), we take c = 3 10 − 5.
55. Suppose that f (x) is continuous on the interval [2, 5] and differentiable on the interval
(2, 5). Show that if 1 ≤ f 0 (x) ≤ 4 for all x ∈ [2, 5], then 3 ≤ f (5) − f (2) ≤ 12.
By the Mean Value Theorem, there exists a number c ∈ (2, 5) such that
f 0 (c) =
f (5) − f (2)
5−2
Since 1 ≤ f 0 (x) ≤ 4 on this interval, we have
1≤
f (5) − f (2)
≤4
3
Multiplying this inequality by 3, we obtain
3 ≤ f (5) − f (2) ≤ 12
34