version 005 – exam review practice problems not for a ... · version 005 – exam review practice...

Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 1

This print-out should have 47 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

Find an equation for the tangent to thegraph of f at the point P (2, f(2)) when

f(x) =5

1− 3x.

1. y +3

5x+

2

5= 0

2. y =3

5x− 11

5correct

3. y =1

5x− 9

5

4. y = 3x− 7

5. y +1

5x+

3

5= 0

Explanation:If x = 2, then f(2) = −1, so we have to find

an equation for the tangent line to the graphof

f(x) =5

1− 3x

at the point (2, −1). Now the Newtonianquotient for f at a general point (x, f(x)) isgiven by

f(x+ h) − f(x)

h.

First let’s compute the numerator of the New-tonian Quotient:

f(x+ h)− f(x) =5

1− 3(x+ h)− 5

1− 3x

=5(1− 3x)− 5{1− 3(x+ h)}

(1− 3x) {1− 3(x+ h)}

=15h

(1− 3h) (1− 3(x+ h)).

Thus

f(x+ h)− f(x)

h=

15

(1− 3x) (1− 3(x+ h)).

Hence

f ′(x) = limh→ 0

15

(1− 3(x+ h))(1− 3x)

=15

(1− 3x)2.

At x = 2, therefore,

f ′(2) =15

(1− 6)2=

3

5,

so by the point slope formula an equation forthe tangent line at (2,−1) is

y + 1 =3

5(x− 2)

which after simplification becomes

y =3

5x− 11

5.

002 10.0 points

Finddy

dxwhen

2√x+

3√y

= 5 .

1.dy

dx=

3

2

(x

y

)3/2

2.dy

dx= −2

3

(y

x

)3/2correct

3.dy

dx=

2

3(xy)1/2

4.dy

dx=

3

2(xy)1/2

5.dy

dx=

2

3

(y

x

)3/2

6.dy

dx= −3

2

(x

y

)3/2

Explanation:


Differentiating implicitly with respect to x,we see that

−1

2

( 2

x√x+

3

y√y

dy

dx

)

= 0 .

Consequently,

dy

dx= −2

3

(y

x

)3/2.

003 10.0 points

Determine the value of dy/dt at x = 3 when

y = x2 − 3x

and dx/dt = 3.

1.dy

dt

∣

∣

∣

x=3= 5

2.dy

dt

∣

∣

∣

x=3= 9 correct

3.dy

dt

∣

∣

∣

x=3= 3

4.dy

dt

∣

∣

∣

x=3= 1

5.dy

dt

∣

∣

∣

x=3= 7

Explanation:Differentiating implicitly with respect to t

we see that

dy

dt= (2x− 3)

dx

dt= 3 (2x− 3) .

At x = 3, therefore,

dy

dt= 3(3) = 9 .

004 10.0 points

Determine f ′(x) when

f(x) =sin(x)− 1

sin(x) + 4.

1. f ′(x) = − 3 cos(x)

(sin(x) + 4)2

2. f ′(x) =5 sin(x) cos(x)

sin(x) + 4

3. f ′(x) = − 5 cos(x)

(sin(x) + 4)2

4. f ′(x) = −3 sin(x) cos(x)

sin(x) + 4

5. f ′(x) =5 cos(x)

(sin(x) + 4)2correct

6. f ′(x) =3 cos(x)

sin(x) + 4

Explanation:By the Quotient Rule,

f ′(x) =(sin(x) + 4) cos(x)− (sin(x)− 1) cos(x)

(sin(x) + 4)2.

But

(sin(x)+4) cos(x)−(sin(x)−1) cos(x) = 5 cos(x) .

Thus

f ′(x) =5 cos(x)

(sin(x) + 4)2.

keywords: derivative of trig functions, deriva-tive, quotient rule

005 10.0 points

The points P and Q on the graph of

y2 − xy − 5y + 10 = 0

have the same x-coordinate x = 2. Find thepoint of intersection of the tangent lines tothe graph at P and Q.

1. intersect at =

(

5

7,20

7

)

correct

2. intersect at =

(

10

3,20

7

)

3. intersect at =

(

5

7, −20

7

)


4. intersect at =

(

20

7,5

3

)

5. intersect at =

(

5

3,20

7

)

Explanation:The respective y-coordinates at P, Q are

the solutions of

(‡) y2 − xy − 5y + 10 = 0

at x = 2; i.e., the solutions of

y2 − 7y + 10 = (y − 5)(y − 2) = 0.

Thus

P = (2, 5), Q = (2, 2).

To determine the tangent lines we need alsothe value of the derivative at P and Q. Butby implicit differentiation,

2ydy

dx− (x+ 5)

dy

dx− y = 0.

so

dy

dx=

y

2y − x− 5.

Thus

dy

dx

∣

∣

∣

P=

5

3,

dy

dx

∣

∣

∣

Q= −2

3.

By the point-slope formula, therefore, theequation of the tangent line at P is

y − 5 =5

3(x− 2),

while that at Q is

y − 2 = −2

3(x− 2).

Consequently, the tangent lines at P and Qare

y =5

3x+

5

3

and

y +2

3x =

10

3

respectively. These two tangent lines

intersect at =

(

5

7,20

7

)

.

006 10.0 points

At noon, ship A is 150 miles due west ofship B. Ship A is sailing south at 30 mphwhile ship B is sailing north at 10 mph.At what speed is the distance between the

ships changing at 5:00 pm?

1. speed = 34 mph

2. speed = 38 mph

3. speed = 32 mph correct

4. speed = 30 mph

5. speed = 36 mph

Explanation:Let a = a(t) be the distance travelled by

ship A up to time t and b = b(t) the distancetravelled by ship B. Then, if s = s(t) isthe distance between the ships, the relativepositions and directions of movement of theships are shown in

150

a

b

s

A

B


By Pythagoras, therefore,

s2 = (a+ b)2 + (150)2 .

Thus, after implicit differentiation,

2sds

dt= 2(a+ b)

( da

dt+

db

dt

)

,

in which case,

ds

dt= (30 + 10)

( a+ b

s

)

= 40( a+ b

s

)

when ships A and B are travelling at respec-tive speeds 30 mph and 10 mph.But at 5:00 pm,

a = 150, b = 50, s = 250

Consequently, at 5:00 pm, the distance be-tween the ships is increasing at a

speed =ds

dt

∣

∣

∣

t=5= 32 mph .

007 10.0 points

Determine dy/dx when

y cos(x2) = 5 .

1.dy

dx= 2xy cos(x2)

2.dy

dx= 2xy tan(x2) correct

3.dy

dx= −2xy tan(x2)

4.dy

dx= 2xy cot(x2)

5.dy

dx= −2xy cot(x2)

6.dy

dx= −2xy sin(x2)

Explanation:

After implicit differentiation with respectto x we see that

−2xy sin(x2) + y′ cos(x2) = 0 .

Consequently,

dy

dx=

2xy sin(x2)

cos(x2)= 2xy tan(x2) .

008 10.0 points

A 5 foot ladder is leaning against a wall. Ifthe foot of the ladder is sliding away from thewall at a rate of 12 ft/sec, at what speed isthe top of the ladder falling when the foot ofthe ladder is 4 feet away from the base of thewall?

1. speed = 16 ft/sec correct

2. speed = 15 ft/sec

3. speed =49

3ft/sec

4. speed =46

3ft/sec

5. speed =47

3ft/sec

Explanation:Let y be the height of the ladder when the

foot of the ladder is x feet from the base ofthe wall as shown in figure

x ft.

5 ft.

We have to express dy/dt in terms of x, y anddx/dt. But by Pythagoras’ theorem,

x2 + y2 = 25 ,


so by implicit differentiation,

2xdx

dt+ 2y

dy

dt= 0 .

In this case

dy

dt= −x

y

dx

dt.

But again by Pythagoras, if x = 4, then y = 3.Thus, if the foot of the ladder is moving awayfrom the wall at a speed of

dx

dt= 12 ft/sec ,

and x = 4, then the velocity of the top of theladder is given by

dy

dt= −4

3

dx

dt.

Consequently, the speed at which the top ofthe ladder is falling is

speed =∣

∣

∣

dy

dt

∣

∣

∣= 16 ft/sec .

keywords: speed, ladder, related rates

009 10.0 points

Find the y-intercept of the tangent line atthe point P (2, f(2)) on the graph of

f(x) = x2 + 3x+ 2 .

1. y-intercept = 1

2. y-intercept = −2 correct

3. y-intercept = 6

4. y-intercept = −6

5. y-intercept = 2

6. y-intercept = −1

Explanation:

The slope, m, of the tangent line at thepoint P (2, f(2)) on the graph of f is thevalue of the derivative

f ′(x) = 2x+ 3

at x = 2, i.e., m = 7. On the other hand,f(2) = 12.

Thus by the point-slope formula, an equa-tion for the tangent line at P (2, f(2)) is

y − 12 = 7(x− 2) ,


y = 7x− 2 .

Consequently, the tangent line at P has

y-intercept = −2 .

010 10.0 points

Determine the derivative of

f(x) = 2 arcsin(x

3

)

.

1. f ′(x) =6√

9− x2

2. f ′(x) =3√

9− x2

3. f ′(x) =2√

1− x2

4. f ′(x) =6√

1− x2

5. f ′(x) =2√

9− x2correct

6. f ′(x) =3√

1− x2

Explanation:Use of

d

dxarcsin(x) =

1√1− x2

,

together with the Chain Rule shows that

f ′(x) =2

√

1− (x/3)2

(

1

3

)

.


Consequently,

f ′(x) =2√

9− x2.

011 10.0 points

Find the derivative of

f(x) = sin−1(e3x) .

1. f ′(x) =3e3x√1− e6x

correct

2. f ′(x) =3√

1− e6x

3. f ′(x) =3e3x

1 + e6x

4. f ′(x) =3

1 + e6x

5. f ′(x) =e3x

1 + e6x

6. f ′(x) =1

1 + e6x

7. f ′(x) =1√

1− e6x

8. f ′(x) =e3x√1− e6x

Explanation:Since

d

dxsin−1 x =

1√1− x2

,d

dxeax = aeax ,

the Chain Rule ensures that

f ′(x) =3e3x√1− e6x

.

012 10.0 points

The height of a triangle is increasing at arate of 4 cm/min while its area is increasingat a rate of 3 sq. cms/min.

At what speed is the base of the trianglechanging when the height of the triangle is 3cms and its area is 15 sq. cms?

1. speed =32

3cms/min

2. speed = 11 cms/min

3. speed = 12 cms/min

4. speed =35

3cms/min

5. speed =34

3cms/min correct

Explanation:Let b be the length of the base and h the

height of the triangle. Then the triangle has

area = A =1

2b h .

Thus by the Product Rule,

dA

dt=

1

2

(

bdh

dt+ h

db

dt

)

,

and so

db

dt=

1

h

(

2dA

dt− b

dh

dt

)

=2

h

( dA

dt− A

h

dh

dt

)

,

since b = 2A/h. Thus, when

dh

dt= 4 , and

dA

dt= 3 ,

we see that

db

dt=

2

h

(

3− 4A

h

)

cms/min .

Consequently, at the moment when

h = 3 and A = 15 ,

the base length is changing at a

speed =34

3cms/min


(recall: speed is non-negative).

013 10.0 points

Use linear approximation with a = 16 toestimate the number

√16.2 as a fraction.

1.√16.2 ≈ 4

1

40correct

2.√16.2 ≈ 4

1

80

3.√16.2 ≈ 4

3

80

4.√16.2 ≈ 4

1

20

5.√16.2 ≈ 4

1

16

Explanation:For a general function f , its linear approxi-

mation at x = a is defined by

L(x) = f(a) + f ′(a)(x− a)

and for values of x near a

f(x) ≈ L(x) = f(a) + f ′(a)(x− a)

provides a reasonable approximation for f(x).

Now set

f(x) =√x, f ′(x) =

1

2√x.

Then, if we can calculate√a easily, the linear

approximation

√a+ h ≈

√a+

h

2√a

provides a very simple method via calculusfor computing a good estimate of the value of√a+ h for small values of h.

In the given example we can thus set

a = 16 , h =2

10.

For then

√16.2 ≈ 4

1

40.

014 10.0 points

Find y′ when

xy + 5x+ 2x2 = 6 .

1. y′ =y + 5 + 2x

x

2. y′ = −y + 5 + 2x

x

3. y′ = −(y + 5 + 4x)

4. y′ =5 + 2x− y

x

5. y′ =y + 5 + 4x

x

6. y′ = −y + 5 + 4x

xcorrect

Explanation:Differentiating implicitly with respect to x

we see that

d

dx

(

xy + 5x+ 2x2)

=d

dx(6) .

Thus

(xy′ + y) + 5 + 4x = 0 ,

and so

xy′ = −y − 5− 4x .

Consequently,

y′ = −y + 5 + 4x

x.

015 10.0 points

Let f, g be differentiable functions. Con-sider the following statements:

A. If F1(x) = {f(x)− 1/f(x)}2, then

F ′

1(x) = 2f ′(x)

{

f(x) +1

f(x)3

}

.


B. If F2(x) = f(x)g(x), then

F ′

2(x) = f ′(x)g(x) + f(x)g′(x) .

C. If F3(x) = f(x)/g(x), then

F ′

3(x) =f ′(x)g(x) + f(x)g′(x)

g(x)2.

Which of these statements are true?

1. C only

2. B and C only

3. A and B only

4. A only

5. all of them

6. B only correct

7. none of them

8. A and C only

Explanation:By the respective Product and Quotient

Rules,

d

dxf(x)g(x) = f ′(x)g(x) + f(x)g′(x)

while

d

dx

f(x)

g(x)=

f ′(x)g(x)− g′(x)f(x)

g(x)2.

Applying these to

F (x) =

(

f(x)− 1

f(x)

)2

= f(x)2 − 2 +1

f(x)2,

we see that

F ′(x) = 2f ′(x)

{

f(x)− 1

f(x)3

}

.

Consequently,

A. Not True

B. True

C. Not True

016 10.0 points

Find the derivative of f when

f(x) =√x(2x+ 7) .

1. f ′(x) =4x+ 7

x√x

2. f ′(x) =6x+ 7

2√x

correct

3. f ′(x) =6x− 7

2√x

4. f ′(x) =4x− 7

x√x

5. f ′(x) =6x− 7

x√x

6. f ′(x) =4x+ 7

2√x

Explanation:By the Product Rule

f ′(x) =2x+ 7

2√x

+ 2√x .

After simplification this becomes

f ′(x) =2x+ 7 + 4x

2√x

=6x+ 7

2√x

.

017 10.0 points


f(x) =(

sin−1(3x))2

.

1. f ′(x) =3√

1− 9x2sin−1(3x)


2. f ′(x) = cos (3x) sin (3x)

3. f ′(x) =6√

9− x2sin−1(3x)

4. f ′(x) = 6 cos (3x) sin (3x)

5. f ′(x) =3√

9− x2sin−1(3x)

6. f ′(x) =6√

1− 9x2sin−1(3x) correct

Explanation:The Chain Rule together with

d

dx

(

sin−1 (ax))

=a√

1− a2x2

shows that

f ′(x) =6√

1− 9x2sin−1(3x) .

018 10.0 points

Find an equation for the tangent line to the

graph of f at the point P(π

4, f(π

4

))

when

f(x) = 9 tan(x) .

1. y = 10x+ 2(

1− π

4

)

2. y = 18x+ 9(

1− π

2

)

correct

3. y = 13x+ 18(

1− π

4

)

4. y = 14x+ 3(

1− π

4

)

5. y = 17x+ 14(

1− π

2

)

Explanation:

When x =π

4, then f(x) = 9, so P =

(π

4, 9)

. Now

f ′(x) = 9 sec2(x) .

Thus the slope of the tangent line at P is

f ′

(π

4

)

= 9 sec2(π

4

)

= 18 .

By the point-slope formula, therefore, anequation for the tangent line at P is givenby

y − 9 = 18(

x− π

4

)

,


y = 18x+ 9(

1− π

2

)

.

019 10.0 points

Find f ′(x) when

f(x) =4x− 1

5x− 1.

1. f ′(x) =5− 4x

(5x− 1)2

2. f ′(x) =1

(5x− 1)2correct

3. f ′(x) =20x− 4

(5x− 1)2

4. f ′(x) =1

5x− 1

5. f ′(x) = − 1

(5x− 1)2

Explanation:Using the Quotient Rule for differentiation

we see that

f ′(x) =4(5x− 1)− 5(4x− 1)

(5x− 1)2.

Consequently,

f ′(x) =1

(5x− 1)2.

020 10.0 points


A circle of radius r has area A and circum-ference C are given respectively by

A = πr2 , C = 2πr .

If r varies with time t, for what value of r isthe rate of change of A with respect to t twicethe rate of change of C with respect to t?

1. r =1

2

2. r = 2π

3. r = 2 correct

4. r = 1

5. r =π

2

6. r = π

Explanation:Differentiating

A = πr2 , C = 2πr

implicitly with respect to t we see that

dA

dt= 2πr

dr

dt,

dC

dt= 2π

dr

dt.

Thus the rate of change, dA/dt, of area istwice the rate of change, dC/dt, of circumfer-ence when

dA

dt= 2

dC

dt,

i.e., when

2πrdr

dt= 2

(

2πdr

dt

)

.

This happens when

r = 2 .

021 10.0 points


f(x) =2− cosx

sinx.

1. f ′(x) =1− 2 cosx

sin2 xcorrect

2. f ′(x) =2 + sinx

cos2 x

3. f ′(x) =1 + 2 cosx

sin2 x

4. f ′(x) =2− sinx

cosx

5. f ′(x) =1 + 2 cosx

sinx

6. f ′(x) =2− sinx

cos2 x

Explanation:By the Quotient Rule,

f ′(x) =(sinx) sinx− (cosx)(2− cosx)

sin2 x

=(cos2 x+ sin2 x)− 2 cosx

sin2 x.

But

cos2 x+ sin2 x = 1 .

Consequently,

f ′(x) =1− 2 cosx

sin2 x.

keywords: DerivTrig, DerivTrigExam, quo-tient rule

022 10.0 points


f(x) = tan−1( x√

3− x2

)

.

(Hint : first simplify f .)

1. f ′(x) =

√3√

3 + x2

2. f ′(x) =x√

x2 − 3


3. f ′(x) =1√

3− x2correct

4. f ′(x) =

√3√

3− x2

5. f ′(x) =x

x2 + 3

Explanation:If

tan θ =x√

3− x2,

then by Pythagoras’ theorem applied to theright triangle

√

3− x2

x

θ

√3

we see thatsin θ =

x√3.

Thusf(x) = sin−1

( x√3

)

.

Consequently,

f ′(x) =1√

3− x2.

Alternatively, we can differentiate f using theChain Rule and the fact that

d

dxtan−1 x =

1

1 + x2.

.

023 10.0 points

Find the derivative of f when

f(x) =tan(x)

8 + 3 sec(x).

1. f ′(x) =sec2(x)

8 + 3 sec(x) tan(x)

2. f ′(x) = − sec2(x)

(8 + 3 sec(x) tan(x))2

3. f ′(x) =8 sec(x) + 3

(8 + 3 sec(x))2

4. f ′(x) =8 cos(x) + 3

(8 cos(x) + 3)2

5. f ′(x) =8 + 3 cos(x)

(8 cos(x) + 3)2correct

6. f ′(x) =8 sin(x) + 3 cos(x)

(8 cos(x) + 3)2

Explanation:Now,

tan(x)

8 + 3 sec(x)=

sin(x)cos(x)

8 + 3cos(x)

=sin(x)

8 cos(x) + 3.

Thus

f ′(x) =cos(x)

8 cos(x) + 3+

8 sin2(x)

(8 cos(x) + 3)2

=cos(x)(8 cos(x) + 3) + 8 sin2(x)

(8 cos(x) + 3)2.

Consequently,

f ′(x) =8 + 3 cos(x)

(8 cos(x) + 3)2,

since cos2(x) + sin2(x) = 1.

024 10.0 points

If y = y(x) is defined implicitly by

6e5y = 3xy + 6x ,

find the value of dy/dx at (x, y) = (1, 0).

1.dy

dx= −2

9

2.dy

dx=

2

9correct

3.dy

dx=

5

27


4.dy

dx= − 7

27

5.dy

dx= − 2

11

6.dy

dx=

2

11


6e5y = 3xy + 6x

implicitly with respect to x we see that

30 e5ydy

dx= 3y + 3x

dy

dx+ 6 .

In this case,

dy

dx

(

30e5y − 3x)

= 3y + 6 .

Consequently, at (x, y) = (1, 0) we see that

dy

dx=

2

9.

keywords: implicit differentiation, exponen-tial function,

025 10.0 points

Boyle’s Law states that when a sample ofgas is compressed at a constant temperature,the pressure and volume satisfy the equationPV = C, where C is a constant. Suppose thatat a certain instant the volume is 400 ccs,the pressure is 80 kPa, and the pressure isincreasing at a rate of 8 kPa/min.At what rate is the volume decreasing at

this instant?

1. rate = 36 ccs/min

2. rate = 40 ccs/min correct




Explanation:After differentiation of PV = C with re-

spect to t using the Product Rule we see that

d(PV )

dt= P

dV

dt+ V

dP

dt= 0 ,

which after rearrangement becomes

dV

dt= −V

P

dP

dt.

When

V = 400 , P = 80 ,dP

dt= 8 ,

therefore,

dV

dt= −400 · 8

80= −40 .

Consequently, the volume is decreasing at a

rate of 40 ccs/min .

026 10.0 points


f(x) = sin−1( x√

2 + x2

)

.

(Hint : first simplify f .)

1. f ′(x) =x√

2 + x2

2. f ′(x) =

√2

2 + x2correct

3. f ′(x) =x

x2 + 2

4. f ′(x) =1

2 + x2

5. f ′(x) =

√2√

2 + x2

Explanation:


Ifsin θ =

x√2 + x2

,

then by Pythagoras’ theorem applied to theright triangle

√2

x

θ

√ 2 +x2

we see that

tan θ =x√2.

Thus

f(x) = θ = tan−1( x√

2

)

.

Consequently,

f ′(x) =

√2

2 + x2.

Alternatively, we can differentiate f using theChain Rule and the fact that

d

dxsin−1 x =

1√1− x2

.

.

027 10.0 points

Find the linearization of

f(x) =1√2 + x

at x = 0.

1. L(x) =1√2

(

1− 1

4x)

correct

2. L(x) =1√2

(

1 +1

4x)

3. L(x) =1

2

(

1 +1

4x)

4. L(x) =1√2− 1

2x

5. L(x) =1√2+

1

2x

6. L(x) =1

2

(

1− 1

2x)

Explanation:The linearization of f is the function

L(x) = f(0) + f ′(0)x .

But for the function

f(x) =1√2 + x

= (2 + x)−1/2 ,

the Chain Rule ensures that

f ′(x) = −1

2(2 + x)−3/2 .

Consequently,

f(0) =1√2, f ′(0) = − 1

4√2,

and so

L(x) =1√2

(

1− 1

4x)

.

028 10.0 points

Find the value of f ′(a) when

f(t) =2t+ 1

t+ 2.

1. f ′(a) =3

a+ 2

2. f ′(a) =3

(a+ 2)2correct

3. f ′(a) = − 4

a + 2

4. f ′(a) = − 4

(a+ 2)2


5. f ′(a) = − 3

(a+ 2)2

6. f ′(a) =4

(a+ 2)2

Explanation:By definition,

f ′(a) = limh→ 0

f(a+ h)− f(a)

h.

Now, for the given f ,

f(a+ h) =2(a+ h) + 1

a+ h+ 2,

while

f(a) =2a+ 1

a+ 2.

Thus

f(a+ h) − f(a) =2(a+ h) + 1

a+ h+ 2− 2a+ 1

a+ 2

=

(

(2(a+ h) + 1)(a+ 2)

− (a+ h+ 2)(2a+ 1)

)

(a+ h+ 2)(a+ 2).

But

{2(a+ h) + 1}(a+ 2)

= 2h(a+ 2) + (2a+ 1)(a+ 2) ,

and

(a+h+2)(2a+1) = h(2a+1)+(a+2)(2a+1) .

Consequently,

f(a+ h)− f(a)

h

=h{2(a+ 2)− (2a+ 1)}h(a+ h+ 2)(a+ 2)

=3

(a+ h+ 2)(a+ 2),

in which case

f ′(a) =3

(a+ 2)2.

029 10.0 points


f(x) = x2 sin(x) + 2x cos(x) .

1. f ′(x) = (x2 − 2) cos(x)

2. f ′(x) = (x2 − 2) sin(x)

3. f ′(x) = (2 + x2) sin(x)

4. f ′(x) = (2− x2) cos(x)

5. f ′(x) = (x2 + 2) cos(x) correct

6. f ′(x) = (2− x2) sin(x)

Explanation:By the Product Rule

d

dx(x2 sin(x)) = 2x sin(x) + x2 cos(x) ,

while

d

dx(2x cos(x)) = 2 cos(x)− 2x sin(x) .

Consquently,

f ′(x) = (x2 + 2) cos(x) .

keywords: DerivTrig, DerivTrigExam,

030 (part 1 of 2) 10.0 points

A point is moving on the graph of

4x3 + 5y3 = xy .

When the point is at

P =(1

9,1

9

)

,

its x-coordinate is decreasing at a speed of 4units per second.


What is the speed of the y-coordinate atthat time?

1. speed y-coord = −3 units/sec

2. speed y-coord = 3 units/sec

3. speed y-coord = −2 units/sec

4. speed y-coord = 1 units/sec

5. speed y-coord = 2 units/sec correct


4x3 + 5y3 = xy


12x2dx

dt+ 15y2

dy

dt= y

dx

dt+ x

dy

dt.

Thusdy

dt=

(

12x2 − y

x− 15y2

)

dx

dt.

Now at P

(

1

9,1

9

)

,

12x2 − y =

(

12

(9)2− 1

9

)

=1

(9)2(3) ,

while

x− 15y2 =

(

1

9− 15

(9)2

)

= − 1

(9)2(6) .

Hence, at P ,

dy

dt= −1

2

dx

dt.

When the x-coordinate at P is decreasing ata rate of 4 units per second, therefore,

dy

dt= 4

(

1

2

)

,

so the

speed y-coord = 2 units/sec .

031 (part 2 of 2) 10.0 pointsIn which direction is the y-coordinate mov-

ing at that time?

1. direction increasing y correct

2. direction decreasing y

Explanation:Since

dy

dt= 2,

at P , the y-coordinate of the point is movingin the

direction increasing y

at that time.

032 10.0 points

Finddy

dxwhen

2x3 + y3 − 9xy − 1 = 0 .

1.dy

dx=

2x2 − 3y

y2 − 3x

2.dy

dx=

2x2 + 3y

y2 − 3x

3.dy

dx=

2x2 − 3y

y2 + 3x

4.dy

dx=

3y − 2x2

y2 − 3xcorrect

5.dy

dx=

3y + 2x2

y2 + 3x

Explanation:We use implicit differentiation. For then

6x2 + 3y2dy

dx− 9y − 9x

dy

dx= 0 ,


which after solving for dy/dx and taking outthe common factor 3 gives

3(

(2x2 − 3y) +dy

dx(y2 − 3x)

)

= 0 .

Consequently,

dy

dx=

3y − 2x2

y2 − 3x.

keywords: implicit differentiation, Folium ofDescartes, derivative,

033 10.0 points

Use linear appproximation to estimate the

value of 171/4. (Hint: (16)1/4 = 2.)

1. 171/4 ≈ 31

16

2. 171/4 ≈ 63

32

3. 171/4 ≈ 33

16

4. 171/4 ≈ 2

5. 171/4 ≈ 65

32correct

Explanation:

Set f(x) = x1/4, so that f(16) = 2 as thehint indicates. Then

df

dx=

1

4x3/4.

By differentials, therefore, we see that

f(a+∆x)− f(a)

≈ df

dx

∣

∣

∣

x= a∆x =

∆x

4a3/4.

Thus, with a = 16 and ∆x = 1,

171/4 − 2 =1

32.

Consequently,

171/4 ≈ 65

32.

034 10.0 points

Find the derivative of g when

g(x) = x4 cos(x) .

1. g′(x) = x3 (4 sin(x)− x cos(x))

2. g′(x) = x3 (4 cos(x)− x sin(x)) correct

3. g′(x) = x3 (4 cos(x) + x sin(x))

4. g′(x) = x3 (4 sin(x) + x cos(x))

5. g′(x) = x4 (3 cos(x)− sin(x))

6. g′(x) = x4 (3 sin(x)− cos(x))

Explanation:By the Product rule,

g′(x) = x4 (− sin(x)) + (cos(x)) · 4x3 .

Consequently,

g′(x) = x3 (4 cos(x)− x sin(x)) .

035 10.0 points

Finddy

dxwhen

tan(x− y) = 3x+ 2y .

1.dy

dx=

sec2(x− y) + 3

sec2(x− y) + 2

2.dy

dx=

2− sec2(x− y)

sec2(x− y)− 3

3.dy

dx=

2 + sec2(x− y)

sec2(x− y) + 3


4.dy

dx=

sec2(x− y)− 3

sec2(x− y)− 2

5.dy

dx=

sec2(x− y)− 3

sec2(x− y) + 2correct

6.dy

dx=

2− sec2(x− y)

sec2(x− y) + 3

Explanation:Differentiating implicitly with respect to x,

we see that

sec2(x− y)(

1− dy

dx

)

= 3 + 2dy

dx.

After rearranging, this becomes

dy

dx

(

sec2(x− y) + 2)

= sec2(x− y)− 3 .

Consequently,

dy

dx=

sec2(x− y)− 3

sec2(x− y) + 2.

keywords:

036 10.0 points

If the radius of a melting snowball decreasesat a rate of 1 ins/min, find the rate at whichthe volume is decreasing when the snowballhas diameter 4 inches.

1. rate = 16π cu.ins/min correct

2. rate = 18π cu.ins/min




Explanation:The volume, V , of a sphere of radius r is

given by

V =4

3πr3 .

Thus by implicit differentiation,

dV

dt= 4πr2

dr

dt= −4πr2 ,

since dr/dt = −1 ins/min. When the snow-ball has diameter 4 inches, therefore, its ra-dius r = 2 and

dV

dt= −4 (4)π .

Consequently, when the snowball has diam-eter 4 inches, the volume of the snowball isdecreasing at a

rate = 16π cu.ins/min .

037 10.0 points

A point is moving on the graph of

6x3 + 4y3 = xy.

When the point is at

P =( 1

10,

1

10

)

,

its y-coordinate is increasing at a speed of 7units per second.

What is the speed of the x-coordinate atthat time and in which direction is the x-coordinate moving?

1. speed =9

4units/sec, increasing x

2. speed =17

8units/sec, decreasing x

3. speed =9


4. speed = 2 units/sec, increasing x

5. speed =7


6. speed = 2 units/sec, decreasing x

7. speed =7


correct


8. speed =17



6x3 + 4y3 = xy


18x2dx

dt+ 12y2

dy

dt= y

dx

dt+ x

dy

dt.

Thus

dx

dt=(x− 12y2

18x2 − y

)dy

dt.

Now at P ,

x− 12y2 = − 1

50,

while

18x2 − y =2

25.

Hence, at P ,

dx

dt= −1

4

dy

dt.

When the y-coordinate at P is increasing at arate of 7 units per second, therefore,

dx

dt= −7

4.

Consequently, the x-coordinate is moving at

speed =7

4units/sec ,

and the negative sign indicates that it is mov-ing in the direction of decreasing x.

038 10.0 points

A rock is thrown into a still pond and causesa circular ripple. If the radius of the ripple isincreasing at a rate of 4 ft/sec, at what speedis the area of the ripple increasing when itsradius is 2 feet?

1. speed = 18π sq. ft/sec


3. speed = 19 sq. ft/sec

4. speed = 16π sq. ft/sec correct





Explanation:The area, A, of a circle having radius r is

given by A = πr2. Differentiating implicitlywith respect to t we thus see that

dA

dt= 2πr

dr

dt.

When

r = 2,dr

dt= 4,

therefore, the speed at which the area of theripple is increasing is given by

speed = 16π sq. ft/sec .

039 10.0 points

Finddy

dxwhen

tan(xy) = x− 3y .

1.dy

dx=

1 + y sec2(xy)

x sec2(xy) + 3

2.dy

dx=

1− y sec2(xy)

x sec2(xy) + 3correct

3.dy

dx=

3 + x sec2(xy)

y sec2(xy)− 1

4.dy

dx=

3− x sec2(xy)

y sec2(xy)− 1


5.dy

dx=

1− y sec2(xy)

x sec2(xy)− 3

6.dy

dx=

3− x sec2(xy)

y sec2(xy) + 1


we see that

sec2(xy)(

y + xdy

dx

)

= 1− 3dy

dx.

After rearranging, this becomes

dy

dx

(

x sec2(xy) + 3)

= 1− y sec2(xy) .

Consequently,

dy

dx=

1− y sec2(xy)

x sec2(xy) + 3.

keywords:

040 10.0 points


g(x) =

(

x+ 2

x+ 3

)

(2x− 3) .

1. g′(x) =2x2 − 12x− 9

x+ 3

2. g′(x) =2x2 + 12x+ 9

x+ 3

3. g′(x) =x2 − 12x+ 9

x+ 3

4. g′(x) =x2 + 12x− 9

(x+ 3)2

5. g′(x) =2x2 + 12x+ 9

(x+ 3)2correct

6. g′(x) =2x2 − 12x− 9

(x+ 3)2

Explanation:

By the Quotient and Product Rules we seethat

g′(x) = 2

(

x+ 2

x+ 3

)

+ (2x− 3)

(

(x+ 3)− (x+ 2)

(x+ 3)2

)

= 2

(

x+ 2

x+ 3

)

+

(

2x− 3

(x+ 3)2

)

=2(x+ 2)(x+ 3) + (2x− 3)

(x+ 3)2.

But

2(x+ 2)(x+ 3) + (2x− 3)

= 2x2 + 12x+ 9 .

Consequently

g′(x) =2x2 + 12x+ 9

(x+ 3)2.

041 10.0 points

Find dy/dx when

3x2 + 2y2 = 5 .

1.dy

dx=

x

2y

2.dy

dx=

3x

2y

3.dy

dx= −3x

2ycorrect

4.dy

dx= −3x

y

5.dy

dx= −3xy

6.dy

dx= 2xy

Explanation:Diferentiating

3x2 + 2y2 = 5


implicitly with respect to x we see that

6x+ 4ydy

dx= 0 .

Consequently,

dy

dx= −6x

4y= −3x

2y.

042 10.0 points

A balloon is released 3 feet away from anobserver. The balloon is rising vertically ata rate of 3 ft/sec and at the same time thewind is carrying it horizontally away from theobserver at a rate of 2 ft/sec. At what speed isthe angle of inclination of the observer’s lineof sight changing 3 seconds after the balloonis released?

1. speed =2

27rads/sec

2. speed =1

27rads/sec

3. speed =5

54rads/sec

4. speed =1

18rads/sec correct

5. speed =1

9rads/sec

Explanation:Let y = y(t) be the height of the balloon t

seconds after release and let x = x(t) be itshorizontal distance from the point of releaseas shown in the figure

3x+ 3

x

y

θ

Then by right triangle trigonometry, the angleof inclination θ satisfies the equation

tan(θ) =y

x+ 3.

Differentiating this equation implicitly withrespect to t we see that

sec2(θ)dθ

dt=

(x+ 3)dy

dt− y

dx

dt(x+ 3)2

.

But sec2(θ) = 1 + tan2(θ). Thus

dθ

dt=

(x+ 3)dy

dt− y

dx

dt(x+ 3)2 + y2

=3(x+ 3)− 2y

(x+ 3)2 + y2rads/sec .

Now after 3 seconds,

x(3) = 3dx

dt= 6 ft/sec ,

while

y(3) = 3dy

dt= 9 ft/sec .

Hence after 3 seconds the speed at which theangle of inclination is changing is given by

speed =1

18rads/sec .

043 10.0 points

The dimensions of a cylinder are changing,but the height is always equal to the diameterof the base of the cylinder. If the height isincreasing at a speed of 5 inches per second,determine the speed at which the volume, V ,is increasing (in cubic inches per second) whenthe height is 2 inches.

1.dV

dt= 14 π cub. ins./sec


2.dV


3.dV


4.dV

dt= 15 π cub. ins./sec correct

5.dV


Explanation:Since the height h of the cylinder is equal

to its diameter D, the radius of the cylinder is

r =1

2h. Thus, as a function of h, the volume

of the cylinder is given by

V (h) = πr2h =π

4h3 .

The rate of change of the volume, therefore, is

dV

dt=

3π

4

(

h2dh

dt

)

.

Now

dh

dt= 5 ins./sec. .

Consequently, when h = 2 ins.,

dV

dt= 15 π cub. ins./sec .

044 10.0 points

Find dy/dx when

y + x = 6√xy .

1.dy

dx=

√

yx − 3

6 +√

xy

2.dy

dx=

3√

yx + 1

1− 3√

xy

3.dy

dx=

3√

yx − 1

1− 3√

xy

correct

4.dy

dx=

3√

yx − 1

1 + 3√

xy

5.dy

dx=

√

yx + 3

6−√

xy

6.dy

dx=

√

yx + 3

6 +√

xy


we see that

dy

dx+ 1 = 3

(√

y

x+

√

x

y

dy

dx

)

,

sody

dx

(

1− 3

√

x

y

)

= 3

√

y

x− 1 .

Consequently,

dy

dx=

3√

yx − 1

1− 3√

xy

.

045 10.0 points

Find an equation for the tangent line to thegraph of

f(x) = sec(x) + 2 cos(x)

at the point (13π, f(13π)).

1. 2y − 3√3x = 1−

√3π

2. 2y + 3√3x =

√3π − 1

3. 2y −√3x = 3− 1√

3π


4. y −√3x = 3− 1√

3π correct

5. y − 3√3x = 1−

√3π

6. y + 3√3x =

√3π − 1

Explanation:Since f(13π) = 3, we have to find an equa-

tion for the tangent line to the graph of f atthe point P (13π, 3). Now

f ′(x) = sec(x) tan(x)− 2 sin(x) ,

so the slope at P is

f ′(13π) =√3 .

By the point-slope formula, therefore, anequation for the tangent line at P is givenby

y − 3 =√3(

x− 1

3π)

,

which after rearrangement and simplificationbecomes

y −√3x = 3− 1√

3π .

046 10.0 points


f(x) =1

3

(

arctan (3x))2

.

1. f ′(x) =1

1 + 9x2arctan (3x)

2. f ′(x) =2

1 + 9x2arctan (3x) correct

3. f ′(x) =1

9 + x2arctan (3x)

4. f ′(x) =2

9 + x2arctan (3x)

5. f ′(x) =1

3sec2(3x) tan(3x)

6. f ′(x) = 2 sec2(3x) tan(3x)

Explanation:Since

d

dxarctan(x) =

1

1 + x2,

the Chain Rule gives

d

dxarctan(3x) =

3

1 + 9x2.

Using the Chain Rule yet again, therefore, wesee that

f ′(x) =2

1 + 9x2arctan (3x) .

keywords: derivative, inverse tan, Chain Rule,

047 10.0 points

A street light is on top of a 12 foot pole. Aperson who is 5 feet tall walks away from thepole at a rate of 3 feet per second. At whatspeed is the length of the person’s shadowgrowing?

1. speed = 2 ft/sec

2. speed =15

7ft/sec correct

3. speed =16

7ft/sec

4. speed =17

7ft/sec

5. speed =18

7ft/sec

Explanation:If x denotes the length of the person’s

shadow and y denotes the distance of theperson from the pole, then shadow and thelightpole are related in the following diagram


12

5

y xx+ y

By similar triangles,

5

x=

12

x+ y,

so 5y = (12− 5)x. Thus, after implicit differ-entiation with respect to t,

5dy

dt= (12− 5)

dx

dt.

When dy/dt = 3, therefore, the length of theperson’s shadow is growing with

speed =15

7ft/sec .

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