– Homework 5 – tran – (52970)
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Multiple-choice questions may continue on
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before answering.
001
1. y
2. y
′
3. y ′
4. y
′
5. y ′
6. y ′
2. f ′ (x) = −12 sin(12x + 5) correct
3. f ′ (x) = 12 sin(12x + 5)
10.0 points
Find the derivative of y when
√
√
√
y = 8 sin( x) − 6 x cos( x) .
′
1
√
√
cos( x)
= 4 sin( x) − √
x
sin(√x) √
√
= 4 sin( x) + 7
x
√
√
sin( x)
= 4 cos( x) − √
x
cos(√x) √
√
= 3 sin( x) − 7
x
√
√
cos( x)
= 3 sin( x) + √
correct
x
sin(√x) √
√
= 3 cos( x) + 7
x
Explanation:
By the Product and Chain Rules,
cos(√x) cos(√x) ′
√
√
y =4
−3
x
x
√x sin(√x) √
+3
.
x
Consequently,
√
√
cos( x)
′
y = 3 sin( x) + √
.
x
002 10.0 points
Find the derivative of f when
f (x) = cos(6x + 8) cos(6x − 3)
− sin(6x + 8) sin(6x − 3) .
1. f ′ (x) = cos(6x + 8) sin(6x − 3)
4. f ′ (x) = −6 sin(12x + 5)
5. f ′ (x) = sin(6x + 8) cos(6x − 3)
Explanation:
Since
f (x) = cos(6x + 8) cos(6x − 3)
− sin(6x + 8) sin(6x − 3)
h
i
= cos (6x + 8) + (6x − 3)
= cos(12x + 5) ,
we see that
f ′ (x) = −12 sin(12x + 5) .
003
10.0 points
Find f ′ (x) when
f (x) = sec2 x + 3 tan2 x .
1. f ′ (x) = −4 sec2 x tan x
2. f ′ (x) = −4 tan2 sec x
3. f ′ (x) = 8 sec2 x tan x correct
4. f ′ (x) = 8 tan2 sec x
5. f ′ (x) = 4 sec2 x tan x
6. f ′ (x) = 4 tan2 sec x
Explanation:
Since
d
sec x = sec x tan x,
dx
d
tan x = sec2 x,
dx
– Homework 5 – tran – (52970)
Thus
the Chain Rule ensures that
f ′ (x) = 2 sec2 x tan x + 6 tan x sec2 x .
Consequently,
f ′ (x) = 8 sec2 x tan x .
004
10.0 points
6
3/2
(x − 3) (x + 3)1/2
4. f ′ (x) =
5. f ′ (x) =
6. f ′ (x) =
3
(x
3)3/2
3
(x −
3)1/2 (x + 3)3/2
correct
3
3/2
(x − 3) (x + 3)1/2
6
1/2
(x − 3) (x + 3)3/2
Explanation:
To apply the Chain Rule it’s simpler to
write
r
1/2
x−3
x−3
f (x) =
=
.
x+3
x+3
For then,
1
f (x) =
2
x−3
x+3
1
=
2
x+3
x−3
−1/2
1/2
1/2
·
6
.
(x + 3)2
Consequently,
3
(x − 3)1/2 (x +
3)3/2
.
10.0 points
f (x) = 5 tan 4x cos3 4x .
1. f ′ (x) = 20 cos 4x (1 + 3 sin2 4x)
2. f ′ (x) = 5 cos 4x(3 sin2 4x − 1)
3. f ′ (x) = 5 cos 4x(1 − 3 cos2 4x)
4. f ′ (x) = 20 cos 4x (1 − 3 cos2 4x)
5. f ′ (x) = 20 cos 4x (1−3 sin2 4x) correct
Explanation:
Using the fact that
d
1
tan x =
,
dx
cos2 x
d x−3
dx x + 3
d x−3
.
dx x + 3
But by the Quotient Rule,
(x + 3) − (x − 3)
d x−3
=
dx x + 3
(x + 3)2
6
=
.
(x + 3)2
d
cos x = − sin x,
dx
together with the Chain rule, we obtain
f ′ (x) =
′
x+3
x−3
Find the derivative of f when
6
1. f (x) = −
3/2
(x − 3) (x + 3)1/2
− 3)1/2 (x +
005
′
3. f ′ (x) = −
1
f (x) =
2
′
f ′ (x) =
Determine f ′ (x) when
r
x−3
f (x) =
.
x+3
2. f ′ (x) =
2
20
cos3 4x
2
cos 4x
− 60 tan 4x cos2 4x sin 4x.
Consequently,
f ′ (x) = 20 cos 4x (1 − 3 sin2 4x) .
Notice that the problem slightly simpler if
we observe that
tan 4x =
sin 4x
,
cos 4x
– Homework 5 – tran – (52970)
3
Determine f ′ (x) when
so that
√
f (x) = 5 sin 4x cos2 4x,
and then differentiate this function using the
known derivatives of sin 4x and cos 4x.
006
10.0 points
′
Find the value of F (2) when
F (x) = f (g(x))
f (x) = e
g(2) = 3,
g (2) = 4 ,
f ′ (2) = 3,
f ′ (3) = 4 .
1. F ′ (2) = 15
.
√
1 e 4x+5
1. f ′ (x) = √
2 4x + 5
√
e
2. f (x) = 2 √
4x+5
′
correct
4x + 5
√
√
3. f ′ (x) = 2e 4x+5 4x + 5
and
′
4x+5
√
4. f ′ (x) = 4e
4x+5
√
4e 4x+5
5. f ′ (x) = √
4x + 5
Explanation:
By the chain rule
′
2. F (2) = 14
√
′
f (x) = e
3. F ′ (2) = 17
4x+5
√
4. F ′ (2) = 13
e
= 2√
d√
4x + 5
dx
4x+5
4x + 5
.
5. F ′ (2) = 16 correct
008
Explanation:
By the Chain Rule,
If y is defined implicitly by
F ′ (x) = f ′ (g(x))g ′(x) .
Thus
′
′
′
F (2) = f (g(2))g (2)
= f ′ (3)g ′ (2) .
Consequently, when
g(2) = 3,
g ′ (2) = 4 ,
f ′ (2) = 3,
f ′ (3) = 4 ,
we see that
F ′ (2) = 16 .
Notice that the value of f ′ (2) was not needed.
007
10.0 points
10.0 points
4y 2 − xy − 15 = 0 ,
find the value of dy/dx at (17, 5).
dy 5
1.
=
correct
dx (17, 5)
23
dy 6
2.
=
dx (17, 5)
23
dy 6
3.
= −
dx (17, 5)
23
dy 5
4.
=
dx (17, 5)
24
5
dy = −
5.
dx (17, 5)
23
Explanation:
– Homework 5 – tran – (52970)
Differentiating implicitly with respect to x
we see that
8y
dy
dy
−y−x
= 0.
dx
dx
Thus
dy
y
=
.
dx
8y − x
keywords: implicit differentiation, Folium of
Descartes, derivative,
010
Find
tan(x + y) = 4x + y .
dy 5
.
=
dx (17, 5)
23
Find
2.
3.
dy
when
dx
dy
2x2 − 3y
= 2
correct
dx
y + 3x
dy
2x2 − 3y
= 2
dx
y − 3x
2x2 + 3y
dy
= 2
dx
y − 3x
dy
3y + 2x2
4.
= 2
dx
y + 3x
5.
3y − 2x2
dy
= 2
dx
y − 3x
Explanation:
We use implicit differentiation. For then
6x2 − 3y 2
1.
dy
4 − sec2 (x + y)
=
correct
dx
sec2 (x + y) − 1
2.
dy
4 + sec2 (x + y)
=
dx
sec2 (x + y) − 1
3.
dy
1 − sec2 (x + y)
=
dx
sec2 (x + y) + 4
4.
1 − sec2 (x + y)
dy
=
dx
sec2 (x + y) − 4
5.
4 − sec2 (x + y)
dy
=
dx
sec2 (x + y) + 1
6.
dy
1 + sec2 (x + y)
=
dx
sec2 (x + y) + 4
10.0 points
2x3 − y 3 − 9xy − 1 = 0 .
1.
10.0 points
dy
when
dx
At (17, 5), therefore,
009
dy
dy
− 9y − 9x
= 0,
dx
dx
which after solving for dy/dx and taking out
the common factor 3 gives
dy
3 (2x2 − 3y) − 3 (y 2 + 3x) = 0 .
dx
Explanation:
Differentiating implicitly with respect to x,
we see that
dy dy
2
= 4+
sec (x + y) 1 +
.
dx
dx
After rearranging, this becomes
dy 2
sec (x + y) − 1 = 4 − sec2 (x + y) .
dx
Consequently,
dy
4 − sec2 (x + y)
=
.
dx
sec2 (x + y) − 1
Consequently,
dy
2x2 − 3y
= 2
dx
y + 3x
4
keywords:
.
011
10.0 points
– Homework 5 – tran – (52970)
After simplification this becomes
y =
013
2. f ′ (x) = √
51
1
.
x−
10
10
3. f ′ (x) = √
10.0 points
4. f ′ (x) = √
Find dy/dx when
e2y = 12x2 + 3y 2 .
1.
dy
4x
=
2
dx
4x + 3y 2 − 3y
2.
dy
3x
=
2
dx
4x + y 2 + y
3.
3x
dy
=
2
dx
4x + y 2 − y
4.
dy
4x
=
correct
2
dx
4x + y 2 − y
5.
4x
dy
=
2
dx
4x + y 2 + y
5. f ′ (x) = √
6. f ′ (x) = √
20
1 − x2
20
16 − x2
5
correct
16 − x2
5
1 − x2
4
1 − x2
Explanation:
Use of
d
1
,
arcsin(x) = √
dx
1 − x2
together with the Chain Rule shows that
5
1
′
f (x) = p
.
2
4
1 − (x/4)
Consequently,
Explanation:
Differentiating
f ′ (x) = √
e2y = 12x2 + 3y 2
implicitly with respect to x we see that
2y dy
dy
2e
= 24x + 6y ,
dx
dx
015
Thus
1. f ′ (x) = √
.
1. f ′ (x) =
2
1 + e6x
2. f ′ (x) =
2e3x
1 + e6x
3. f ′ (x) = √
10.0 points
Determine the derivative of
x
f (x) = 5 arcsin
.
4
4
16 − x2
10.0 points
f (x) = 2 tan−1 (e3x ) .
dy
12x
= 2y
.
dx
e − 3y
014
5
.
16 − x2
Find the derivative of
so
dy
4x
=
2
dx
4x + y 2 − y
6
4. f ′ (x) = √
5. f ′ (x) =
6
1 − e6x
6e3x
1 − e6x
6
1 + e6x
6. f ′ (x) = √
2e3x
1 − e6x
– Homework 5 – tran – (52970)
2
7. f (x) = √
1 − e6x
′
6e3x
correct
1 + e6x
8. f ′ (x) =
Explanation:
Since
1
d
tan−1 x =
,
dx
1 + x2
d ax
e = aeax ,
dx
the Chain Rule ensures that
6e3x
f ′ (x) =
.
1 + e6x
016
10.0 points
Find the derivative of f when
f (θ) = ln (sin 6θ) .
1
cos 6θ
1. f ′ (θ) =
2. f ′ (θ) = 6 cot 6θ correct
Find the derivative of f when
p
f (x) = 3 ln(x + x2 − 7), (x > 3) .
1. f ′ (x) = − √
2. f ′ (x) = − √
6
x2 − 7
3
x2 − 7
3
3. f ′ (x) = − √
2 x2 − 7
4. f ′ (x) =
3
2 x2 − 7
√
5. f ′ (x) = √
6. f ′ (x) = √
3
x2 − 7
x2 − 7
Explanation:
By the Chain Rule
3
x
√
f (x) =
1+ √
x + x2 − 7
x2 − 7
′
= √
018
6
sin 6θ
5. f ′ (θ) =
6. f ′ (θ) = − tan 6θ
Explanation:
By the Chain Rule,
f ′ (θ) =
1
d
6 cos 6θ
(sin 6θ) =
.
sin(6θ) dθ
sin 6θ
3
x2
f (θ) = 6 cot 6θ .
017
10.0 points
−7
f (x) = ln
1. f ′ (x) = −
2x3
1 − x4
2. f ′ (x) = −
2x
1 − x4
3. f ′ (x) =
1 + x2
.
1 − x2
2x
correct
1 − x4
4. f ′ (x) = −
5. f ′ (x) =
.
10.0 points
Find the derivative of
s
Consequently,
′
correct
6
3. f ′ (θ) = cot 6θ
4. f ′ (θ) = 6 tan 6θ
7
x3
1 − x2
2x3
1 − x4
– Homework 5 – tran – (52970)
x
1 − x2
6. f ′ (x) =
8. y ′ = −y(ln x + 1)
Explanation:
Properties of logs ensure that
s
1 + x2
ln
1 − x2
q
q
= ln (1 + x2 ) − ln (1 − x2 )
=
By the Chain rule, therefore,
2x 1 2x
+
f ′ (x) =
2 1 + x2 1 − x2
Consequently,
(1 − x2 ) + (1 + x2 ) 1 − x2
2x
f (x) =
1 − x4
′
019
.
Determine y when
2
y = x(−x ) .
y
(ln x − 1)
x2
2. y ′ = xy(2 ln x + 1)
3. y ′ = −y(2 ln x − 1)
4. y ′ = −
y
(ln x − 1)
x2
5. y ′ = y(ln x + 1)
6. y ′ = −xy(2 ln x + 1) correct
′
ln y = −x2 ln x .
Thus by implicit differentiation,
1 dy
= −2x ln x − x .
y dx
Consequently,
y ′ = −xy(2 ln x + 1) .
020
Find
7. y = xy(2 ln x − 1)
10.0 points
dy
when
dx
y ln x + y 2 = 4x .
.
1.
dy
4 + xy
=
dx
x ln x − 2y
2.
4−y
dy
=
dx
ln x + 2y
3.
4x − y
dy
=
correct
dx
x ln x + 2xy
4.
dy
4+y
=
dx
ln x + 2y
5.
4x − y
dy
=
dx
ln x − 2xy
10.0 points
′
1. y ′ =
Explanation:
After taking natural logs we see that
1
ln(1 + x2 ) − ln(1 − x2 ) .
2
= x
8
6.
dy
4x + y
=
dx
x ln x − 2xy
Explanation:
Differentiating implicitly with respect to x
we see that
1
dy
dy
y
+ (ln x)
+ 2y
= 4,
x
dx
dx
which after rearrangement becomes
dy
y
(ln x + 2y)
= 4− .
dx
x
Consequently,
dy
4x − y
=
dx
x ln x + 2xy
.