– 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) = −4 sin(4x + 4) correct
3. f ′ (x) = sin(2x + 8) cos(2x − 4)
10.0 points
Find the derivative of y when
√
√
√
y = 6 sin( x) − 2 x cos( x) .
′
1
sin(√x) √
cos( x) + 4
x
sin(√x) √
√
3 sin( x) + 4
x
sin(√x) √
√
3 cos( x) − 2
x
√
cos( x) √
√
sin( x) − 4
x
cos(√x) √
√
3 sin( x) − 2
x
cos(√x) √
√
correct
sin( x) + 2
x
√
Explanation:
By the Product and Chain Rules,
cos(√x) cos(√x)
′
√
y =3
− √
x
x
√
√
x sin( x)
√
+
.
x
Consequently,
cos(√x) √
′
√
y = sin( x) + 2
.
x
002 10.0 points
Find the derivative of f when
f (x) = cos(2x + 8) cos(2x − 4)
− sin(2x + 8) sin(2x − 4) .
1. f ′ (x) = cos(2x + 8) sin(2x − 4)
4. f ′ (x) = 4 sin(4x + 4)
5. f ′ (x) = −2 sin(4x + 4)
Explanation:
Since
f (x) = cos(2x + 8) cos(2x − 4)
− sin(2x + 8) sin(2x − 4)
h
i
= cos (2x + 8) + (2x − 4)
= cos(4x + 4) ,
we see that
f ′ (x) = −4 sin(4x + 4) .
003
10.0 points
Find f ′ (x) when
f (x) = 2 sec2 x − 3 tan2 x .
1. f ′ (x) = 2 tan2 sec x
2. f ′ (x) = 10 tan2 sec x
3. f ′ (x) = 2 sec2 x tan x
4. f ′ (x) = −2 sec2 x tan x correct
5. f ′ (x) = 10 sec2 x tan x
6. f ′ (x) = −2 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) = 4 sec2 x tan x − 6 tan x sec2 x .
Consequently,
f ′ (x) = −2 sec2 x tan x .
004
10.0 points
(x −
5. f ′ (x) = −
6. f ′ (x) = −
correct
6
3/2
(x − 3) (x + 3)1/2
3
1/2
(x − 3) (x + 3)3/2
For then,
Consequently,
x−3
x+3
1
=
2
x+3
x−3
−1/2
1/2
3
(x − 3)1/2 (x +
3)3/2
.
10.0 points
3. f ′ (x) = 3 cos 5x(3 sin2 5x − 1)
4. f ′ (x) = 15 cos 5x (1−3 sin2 5x) correct
5. f ′ (x) = 15 cos 5x (1 − 3 cos2 5x)
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) =
1
f (x) =
2
6
.
(x + 3)2
2. f ′ (x) = 15 cos 5x (1 + 3 sin2 5x)
Explanation:
To apply the Chain Rule it’s simpler to
write
r
1/2
x−3
x−3
f (x) =
=
.
x+3
x+3
′
·
1. f ′ (x) = 3 cos 5x(1 − 3 cos2 5x)
3
3/2
(x − 3) (x + 3)1/2
3
1/2
f (x) = 3 tan 5x cos3 5x .
6
1/2
(x − 3) (x + 3)3/2
3)1/2 (x + 3)3/2
x+3
x−3
Find the derivative of f when
6
1. f (x) =
3/2
(x − 3) (x + 3)1/2
4. f ′ (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
15
cos3 5x
2
cos 5x
− 45 tan 5x cos2 5x sin 5x.
Consequently,
f ′ (x) = 15 cos 5x (1 − 3 sin2 5x) .
Notice that the problem slightly simpler if
we observe that
tan 5x =
sin 5x
,
cos 5x
– Homework 5 – tran – (52970)
3
Determine f ′ (x) when
so that
√
f (x) = 3 sin 5x cos2 5x,
and then differentiate this function using the
known derivatives of sin 5x and cos 5x.
006
10.0 points
Find the value of F ′ (6) when
F (x) = f (g(x))
and
g(6) = 2,
g ′ (6) = 2 ,
f ′ (6) = 3,
f ′ (2) = 6 .
1. F ′ (6) = 15
f (x) = e
2x+3
.
√
2e 2x+3
1. f (x) = √
2x + 3
′
√
2. f ′ (x) = 2e
√
e
3. f (x) = √
′
2x+3
2x+3
2x + 3
correct
√
1 e 2x+3
4. f ′ (x) = √
2 2x + 3
√
√
5. f ′ (x) = e 2x+3 2x + 3
Explanation:
By the chain rule
2. F ′ (6) = 13
√
′
f (x) = e
3. F ′ (6) = 12 correct
2x+3
√
e
= √
′
4. F (6) = 14
d√
2x + 3
dx
2x+3
2x + 3
.
5. F ′ (6) = 16
008
Explanation:
By the Chain Rule,
If y is defined implicitly by
F ′ (x) = f ′ (g(x))g ′(x) .
Thus
′
′
′
F (6) = f (g(6))g (6)
= f ′ (2)g ′ (6) .
Consequently, when
g(6) = 2,
g ′ (6) = 2 ,
f ′ (6) = 3,
f ′ (2) = 6 ,
we see that
F ′ (6) = 12 .
Notice that the value of f ′ (6) was not needed.
007
10.0 points
10.0 points
2y 2 − xy − 12 = 0 ,
find the value of dy/dx at (10, 6).
dy 1
1.
= −
dx (10, 6)
2
dy 3
= −
2.
dx (10, 6)
7
dy 1
3.
=
dx (10, 6)
2
dy 3
4.
=
correct
dx (10, 6)
7
2
dy =
5.
dx (10, 6)
5
Explanation:
– Homework 5 – tran – (52970)
Differentiating implicitly with respect to x
we see that
4y
dy
dy
−y−x
= 0.
dx
dx
Thus
dy
y
=
.
dx
4y − x
keywords: implicit differentiation, Folium of
Descartes, derivative,
010
Find
tan(x + y) = 3x + y .
dy 3
.
=
dx (10, 6)
7
1.
dy
1 + sec2 (x + y)
=
dx
sec2 (x + y) + 3
2.
dy
1 − sec2 (x + y)
=
dx
sec2 (x + y) + 3
3.
dy
3 − sec2 (x + y)
=
correct
dx
sec2 (x + y) − 1
4.
3 + sec2 (x + y)
dy
=
dx
sec2 (x + y) − 1
dy
3y + 2x2
= 2
dx
y + 3x
5.
1 − sec2 (x + y)
dy
=
dx
sec2 (x + y) − 3
3y − 2x2
dy
= 2
correct
dx
y − 3x
6.
dy
3 − sec2 (x + y)
=
dx
sec2 (x + y) + 1
Find
10.0 points
dy
when
dx
2x3 + y 3 − 9xy − 1 = 0 .
1.
2.
3.
dy
2x2 − 3y
= 2
dx
y − 3x
dy
2x2 + 3y
4.
= 2
dx
y − 3x
5.
10.0 points
dy
when
dx
At (10, 6), therefore,
009
2x2 − 3y
dy
= 2
dx
y + 3x
Explanation:
We use implicit differentiation. For then
6x2 + 3y 2
dy
dy
− 9y − 9x
= 0,
dx
dx
which after solving for dy/dx and taking out
the common factor 3 gives
dy 2
3 (2x2 − 3y) +
(y − 3x) = 0 .
dx
Explanation:
Differentiating implicitly with respect to x,
we see that
dy dy
2
= 3+
sec (x + y) 1 +
.
dx
dx
After rearranging, this becomes
dy 2
sec (x + y) − 1 = 3 − sec2 (x + y) .
dx
Consequently,
dy
3 − sec2 (x + y)
=
.
dx
sec2 (x + y) − 1
Consequently,
dy
3y − 2x2
= 2
dx
y − 3x
4
keywords:
.
011
10.0 points
– Homework 5 – tran – (52970)
After simplification this becomes
y =
013
2. f ′ (x) = √
9
1
.
x−
4
4
3. f ′ (x) = √
10.0 points
4. f ′ (x) = √
Find dy/dx when
e2y = 21x2 + 7y 2 .
1.
dy
3x
=
2
dx
3x + 7y 2 − 7y
2.
dy
7x
=
2
dx
3x + y 2 − y
3.
3x
dy
=
correct
2
dx
3x + y 2 − y
4.
dy
3x
=
2
dx
3x + y 2 + y
5.
7x
dy
=
2
dx
3x + y 2 + y
5. f ′ (x) = √
6. f ′ (x) = √
6
2
1 − x2
6
1 − x2
3
correct
4 − x2
3
1 − x2
2
4 − x2
Explanation:
Use of
1
d
arcsin(x) = √
,
dx
1 − x2
together with the Chain Rule shows that
3
1
′
f (x) = p
.
1 − (x/2)2 2
Consequently,
Explanation:
Differentiating
f ′ (x) = √
e2y = 21x2 + 7y 2
implicitly with respect to x we see that
2y dy
dy
2e
= 42x + 14y ,
dx
dx
015
3
.
4 − x2
10.0 points
Find the derivative of
f (x) = tan−1 (e3x ) .
so
1. f ′ (x) = √
dy
21x
= 2y
.
dx
e − 7y
2. f ′ (x) =
Thus
dy
3x
=
2
dx
3x + y 2 − y
014
Determine the derivative of
x
f (x) = 3 arcsin
.
2
1. f ′ (x) = √
6
4 − x2
3e3x
correct
1 + e6x
.
3. f ′ (x) = √
10.0 points
4. f ′ (x) =
1
1 − e6x
3
1 + e6x
5. f ′ (x) = √
6. f ′ (x) =
e3x
1 − e6x
3e3x
1 − e6x
1
1 + e6x
– Homework 5 – tran – (52970)
Find the derivative of f when
p
f (x) = 9 ln(x + x2 − 4), (x > 3) .
e3x
7. f ′ (x) =
1 + e6x
8. f ′ (x) = √
3
1 − e6x
1. f ′ (x) = √
Explanation:
Since
d
1
tan−1 x =
,
dx
1 + x2
d ax
e = aeax ,
dx
the Chain Rule ensures that
3e3x
f ′ (x) =
.
1 + e6x
016
10.0 points
Find the derivative of f when
f (θ) = ln (sin 2θ) .
1. f ′ (θ) = 2 cot 2θ correct
2. f ′ (x) = √
1
cos 2θ
3. f ′ (θ) =
′
4. f (θ) = cot 2θ
5. f ′ (θ) = − tan 2θ
4. f ′ (x) =
f ′ (θ) =
1
d
2 cos 2θ
(sin 2θ) =
.
sin(2θ) dθ
sin 2θ
Consequently,
f ′ (θ) = 2 cot 2θ .
017
10.0 points
9
x2 − 4
correct
9
√
2 x2 − 4
5. f ′ (x) = − √
6. f ′ (x) = − √
9
x2 − 4
18
x2 − 4
Explanation:
By the Chain Rule
9
x
√
1+ √
x + x2 − 4
x2 − 4
f ′ (x) =
= √
018
9
x2 − 4
f (x) = ln
1. f ′ (x) =
4x3
1 − 4x4
2x
1 − 4x2
4. f ′ (x) = −
5. f ′ (x) =
1 + 2x2
.
1 − 2x2
4x
correct
1 − 4x4
2. f ′ (x) = −
3. f ′ (x) =
.
10.0 points
Find the derivative of
s
6. f ′ (θ) = 2 tan 2θ
Explanation:
By the Chain Rule,
18
x2 − 4
9
3. f ′ (x) = − √
2 x2 − 4
2
sin 2θ
2. f ′ (θ) =
7
4x
1 − 4x4
4x3
1 − 4x4
– Homework 5 – tran – (52970)
6. f ′ (x) = −
2x3
1 − 4x2
8. y ′ = xy(2 ln x + 1) correct
Explanation:
Properties of logs ensure that
s
1 + 2x2
ln
1 − 2x2
q
q
= ln (1 + 2x2 ) − ln (1 − 2x2 )
=
By the Chain rule, therefore,
4x 1 4x
+
f ′ (x) =
2 1 + 2x2 1 − 2x2
(1 − 2x2 ) + (1 + 2x2 ) 1 − 4x2
Consequently,
4x
f (x) =
1 − 4x4
′
019
ln y = x2 ln x .
Thus by implicit differentiation,
1 dy
= 2x ln x + x .
y dx
Consequently,
Determine y when
2
y = x(x ) .
1. y ′ = y(ln x + 1)
2. y ′ = −xy(2 ln x + 1)
3. y ′ = −
.
y ′ = xy(2 ln x + 1) .
020
Find
y
(ln x − 1)
x2
4. y ′ = −y(2 ln x − 1)
5. y ′ = −y(ln x + 1)
6. y ′ = xy(2 ln x − 1)
y
7. y = 2 (ln x − 1)
x
10.0 points
dy
when
dx
y ln x − y 2 = 3x .
.
1.
dy
3x − y
=
correct
dx
x ln x − 2xy
2.
3x + y
dy
=
dx
x ln x + 2xy
3.
3+y
dy
=
dx
ln x − 2y
10.0 points
′
′
Explanation:
After taking natural logs we see that
1
ln(1 + 2x2 ) − ln(1 − 2x2 ) .
2
= 2x
8
4.
dy
3 + xy
=
dx
x ln x + 2y
5.
3x − y
dy
=
dx
ln x + 2xy
6.
dy
3−y
=
dx
ln x − 2y
Explanation:
Differentiating implicitly with respect to x
we see that
1
dy
dy
y
+ (ln x)
− 2y
= 3,
x
dx
dx
which after rearrangement becomes
dy
y
(ln x − 2y)
= 3− .
dx
x
Consequently,
dy
3x − y
=
dx
x ln x − 2xy
.