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MAT 167 TEST 2 SOLUTIONS
1. FROM FORM A
3. For each function or equation, compute the derivative.
(a) y = 10 cos x − 2 arctan x
(c) g ( u ) =
q

11 u 2 + 1
(b) G ( y ) =
y2
y +1
4
(d) p ( x ) = x ln x
(e) x 2 + y 2 = 2x 2 + 2y 2 − x
Solutions:
(a) y 0 = −10 sin x −

2
2
1+ x 2
(b) Observe that G is a composition of functions; in particular, G ( u ) = u 4 where u =
G0
= G0
· u 0.
y2
.
y +1
By
the Chain Rule, ( y )
(u )
0
• To compute G ( u ), use the Power Rule and G 0 ( u ) = 4u 3 .
• To compute u 0 , use the Quotient Rule and
u0 =
• Thus
2y · ( y + 1) − y 2 · 1
( y + 1)2


G 0 ( y ) = 4
=
y 2 + 2y
( y + 1)2
.
!3  2
 y + 2y
.
·
y +1
( y + 1)2
y2
Remark. We do not need implicit differentiation here. G is a function of y, so the derivative
dy
is with respect to y, not with respect to x. Thus d x does not need to appear in the derivative.
p
(c) Observe that g is a composition of functions; in particular, g ( v ) = 11v where v =
u 2 + 1. (There are other ways to describe this composition.) By the Chain Rule, g 0 ( u ) =
g 0 (v ) · v 0.
p
p
• To compute g 0 ( v ), notice that g ( v ) contains a constant multiple: g ( v ) = 11 · p v.
p
1
So use the Power Rule and the Constant Multiple Rule and g 0 ( v ) = 11 · 12 v − 2 = 2p11v .
• To compute v 0 , use the Sum Rule, the Power Rule, and the Constant Rule and v 0 = 2u.
• Thus
p
11
g 0 (u ) = p
· 2u.
2 u2 + 1
Remark. Notice that if u is already used and you need to identify a composition, you can
always use another variable.
1
(d) To compute p 0 ( x ), use the Product Rule and p 0 ( x ) = 1 · ln x + x · x1 = ln x + 1.
(e) To compute y 0 , we need implicit differentiation. (Here y depends on x.) So

2 i
d 2
x + y 2 = 2x 2 + 2y 2 − x
dx
Right-hand side: Chain Rule!
Let u = 2x 2 + 2y 2 − x;
then u 0 = 4x + 4y · y 0 − 1.

2x + 2y · y 0 = 2 2x 2 + 2y 2 − x · 4x + 4y · y 0 − 1
2x + 2y · y 0 = 16x 3 + 16x 2 y · y 0 − 4x 2
+ 16xy 2 + 16y 3 · y 0 − 4y 2
− 8x 2 − 8xy · y 0 + 2x

2x + 2y · y 0 = 16x 3 + 16xy 2 − 12x 2 − 4y 2 + 2x

+ 16x 2 y + 16y 3 − 8xy · y 0

−
2x
2x
2y − 16x 2 y + 16y 3 − 8xy · y 0 = 16x 3 + 16xy 2 − 12x 2 − 4y 2 + 16x 3 + 16xy 2 − 12x 2 − 4y 2
y =

.
2y − 16x 2 y + 16y 3 − 8xy
0
4. Suppose the equation of a shock absorber is modeled by the equation s ( t ) = 2e −0.9 sin (2πt ),
where s is measured in centimeters and t in seconds.
(a) Find the velocity, v, after t seconds.
(b) What are the units of the velocity?
(c) Is the shock absorber compressing or expanding at time t = 3?
Solutions:
(a) v ( t ) = s 0 ( t ) = 2e −0.9 cos (2πt ) · 2π.
Notice that e −0.9 is a constant (t does not appear in the exponent!) so we use the Constant
Multiple Rule, along with the Chain Rule with u = 2πt because cos (2πt ) is a composition
of functions.
(b) The units of velocity are cm/sec (change in s per change in t ).
(c) At t = 3, v ( t ) = 2e −0.9 cos (2π · 3) · 2π = 2e −0.9 cos (6π) = 2e −0.9 · 1 > 0 so the absorber is
expanding.
5. Differentiate
7
x 12 − 2x 8 + 3
y=
5
3 .
2
3
x −1
x + x +2
Hint: Use logarithmic differentiation.
Solution:
Using properties of logarithms, we have


7
12
8


x − 2x + 3

ln y = ln 

 2
5
3
3
x −1
x + x +2

5
3
7
x3 + x + 2
= ln x 12 − 2x 8 + 3 − ln x 2 − 1

3 i
5

7 h
3
2
12
8
= ln x − 2x + 3 − ln x − 1 + ln x + x + 2

= 7 ln x 12 − 2x 8 + 3 − 5 ln x 2 − 1 + 3 ln x 3 + x + 2 .
Now differentiate both sides with respect to x, and we have (dont’ forget the Chain Rule!)
1
y
· y0 = 7 ·
1

· 12x 11 − 16x 7
x 12 − 2x 8 + 3

1
1
2
− 5· 2
· 2x + 3 · 3
· 3x + 1
x −1
x + x +2
12 − 2x 8 + 3 7

x
1
0
11
7
·
7
·
y =
·
12x
−
16x
5
3
x 12 − 2x 8 + 3
x2 − 1
x3 + x + 2

1
1
2
· 2x + 3 · 3
· 3x + 1
− 5· 2
.
x −1
x + x +2
2. FROM FORM B
2. A mass on a spring vibrates according to the equation x ( t ) = 4 cos (3π ( t + 2)), where t is
measured in seconds and x is measured in centimeters.
(a) Find the velocity, v, and the acceleration, a, at time t .
(b) What are the units of the velocity and the acceleration?
(c) In what direction is the mass moving at time t = 12?
(d) At what time(s) is the mass stationary? (That is, when is its velocity zero?)
Solutions:
(a) Use the Constant Multiple Rule and the Chain Rule (u = 3π · ( t + 2)) to compute
v ( t ) = x 0 ( t ) = 4 · [− sin (3π ( t + 2)) · 3π]
and then
a ( t ) = v 0 ( t ) = −12π cos (3π ( t + 2)) · 3π.
(b) The units of velocity are cm/sec. The units of acceleration are cm2 /sec.
(c) At t = 12,
v (12) = −12π sin (3π (12 + 2)) = 4 · 0 = 0,
so the mass is not moving at time t = 12.
(d) The mass is stationary when v ( t ) = 0. Thus we need to solve
−12π sin (3π ( t + 2)) = 0
sin (3π ( t + 2)) = 0
3π ( t + 2) = πk, k ∈ Z
t +2 =
k
3
, k ∈Z
k
t = −2 + , k ∈ Z.
3
3. For each function or equation, compute the derivative.
p
p
(a) g ( u ) = 2u − 5u
(c) y =
p
(b) p ( x ) = e 2x sin (2x )
(d) G ( y ) =
sin (3x )
2+sin y
2y +cos y
(e) y sin x 2 = x sin y 2
Solutions:
p p
p
(a) Observe that we have two constant multiples: g ( u ) = 2 · u − 5 · u. (Also, u is not in
the second square root. Pay attention! This was a WebAssign problem!) Using the Constant
Multiple and Power Rules,
p
p
p
1
2 p
1
g 0 ( u ) = 2 · u − 2 − 5 · 1 = p − 5.
2
2 u
(b) Observe that p ( x ) is a composition of functions: p ( u ) = e u sin u where u = 2x. By the
Chain Rule, p 0 ( x ) = p 0 ( u ) · u 0 .
• To compute p 0 ( u ), use the Product Rule and p 0 ( u ) = e u sin u + e u cos u.
• To compute u 0 , use the Constant MultipleRule and u 0 = 2 · 1 = 2.
• Thus p 0 ( x ) = e 2x sin (2x ) + e 2x cos (2x ) · 2.
p
(c) Observe that y ( x ) is a double composition of functions: y ( v ) = v where v ( u ) = sin u
and u ( x ) = 3x. By the Chain Rule,
y 0 ( x ) = y 0 (v ) · v 0 ( u ) · u 0 ( x ) .
1
• To compute y 0 ( v ), use the Power Rule and y 0 ( v ) = 12 v − 2 =
1
p
.
2 v
• You should know that v 0 ( u ) = cos u.
• To compute u 0 ( x ), use the Constant Multiple Rule and u 0 ( x ) = 3 · 1 = 3.
• Thus
1
3 cos (3x )
y0 (x ) = p
· cos (3x ) · 3 = p
.
2 sin (3x )
2 sin (3x )
(d) To compute G 0 ( y ), use the Quotient Rule and
G 0 (y ) =
(0 + cos y ) · (2y + cos y ) − (2 + sin y ) · (2 − sin y )
(2y + cos y )2
=
*1
2 y − 4 + 2y cos y + cos
sin2 y
(2y + cos y )2
2y cos y − 3
=
.
(2y + cos y )2
Remark. We do not need implicit differentiation here. G is a function of y, so the derivative
dy
is with respect to y, not with respect to x. Thus d x does not need to appear in the derivative.
(e) To compute y 0 , we need implicit differentiation. (Here y depends on x.) So
d
dx

y sin x 2 = x sin y 2
Both sides: Product Rule!
Left hand side: Chain Rule!
2
Let u = x ;
Right hand side: Chain Rule!
Let u = y 2 ;
then u 0 = 2x.
then u 0 = 2y · y 0 .

y 0 · sin x 2 + y · cos x 2 · 2x = 1 · sin y 2 + x · cos y 2 · 2y · y 0

y 0 sin x 2 − 2xy cos y 2 = sin y 2 − 2xy cos x 2
2 − 2xy cos x 2
sin
y
y0 =

.
sin x 2 − 2xy cos y 2
5. Differentiate

arctan x
y = x2 + 1
.
Hint: Use logarithmic differentiation.
Solution:
Using properties of logarithms, we have
arctan x x2 + 1

ln y = arctan x · ln x 2 + 1 .
ln y = ln

Now differentiate both sides with respect to x, and we have (dont’ forget the Product and Chain
Rules!)

1
1 0
1
· ln x 2 + 1 + arctan x · 2
· 2x
·y = 2
y
x +1
x +1

1 0
1
·y = 2
· ln x 2 + 1 + 2x arctan x
y
x +1

arctan x

1
· 2
y0 = x2 + 1
· ln x 2 + 1 + 2x arctan x
x +1

arctan x−1
· ln x 2 + 1 + 2x arctan x .
y0 = x2 + 1

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