Math 1100-6
October 29, 2008
Math 1100: Assignment #4 SOLUTIONS
Due: Friday Oct 24, 2008 @ 10:30am
No late assignments will be accepted
Please complete the following questions for the assignment. Please be mindful of “good
presentation” criteria outlined in the syllabus as you’re preparing to hand it in.
§11.1
6.
f (x) = ln(x3 ) = 3 ln x
3
f 0 (x) =
x
10.
y = ln(8x3 − 2x) − 2x = ln(2x) + ln(x + 1) + ln(x − 1) − 2x
1
1
1
y0 =
+
+
−2
x x+1 x−1
17. a)
y = ln(4x − 1) − 3 ln(x)
3
4
−
y0 =
4x − 1 x
b) Note that
µ
y = ln
4x − 1
x3
¶
= ln(4x − 1) − ln(x3 )
= ln(4x − 1) − 3 ln x
Therefore, the function in b) is the same as in a).
20.
£
¤
s = ln t3 (t2 − 1)
= ln(t3 ) + ln(t2 − 1)
→
s0
= 3 ln t + ln(t − 1) + ln(t + 1)
3
1
1
=
+
+
t
t−1 t+1
28.
1 + ln x
x2
1
· x2 − (1 + ln x)(2x)
Using the quotient rule
= x
x4
1 − (1 + ln x)2
=
Dividing out a common factor of x
x3
1 + 2 ln x
= −
x3
y =
y0
Page 1
Math 1100-6
October 29, 2008
Page 2
34.
p
ln(3x + 1)
1
1
1
=
·p
·
·3
2
3x
+1
ln(3x + 1)
3
p
=
2(3x + 1) ln(3x + 1)
y =
y0
37.
1
ln(x4 − 4x3 + 1)
ln 6
y = log6 (x4 − 4x3 + 1) =
→
y0 =
=
1
1
·
(4x3 − 12x2 )
ln 6 x4 − 4x3 + 1
4x2 (x − 3)
ln 6(x4 − 4x3 + 1)
43.
C(x) = 1500 + 200 ln(2x + 1)
a)
M C(x) := C 0 (x) =
400
2x + 1
400
≈ 0.9975. This means that the cost will increase by approx $1.00 if one
b) C 0 (200) =
401
more unit is produced.
c) Because cost is always increasing, the marginal cost function must always be positive
(for x ≥ 0). This is true about the problem above.
46.
p = 10 + 50 ln(3x + 1)
a)
p0 =
150
3x + 1
150
= 1.5
100
c) The approximate price increase from 33 to 34 units produced is $1.50 (from part b).
b) p0 (33) =
§11.2
2.
y = x2 − 3 ex
y 0 = 2x − 3 ex
Math 1100-6
October 29, 2008
Page 3
10.
√
y =
y0 =
e
x2 −9
√
x
2
√
e x −9
x2 − 9
18.
p = 4q eq
3
3
3
p0 = 4 eq + 4q eq 3q 2
3
= 4 eq (1 + 3q 3 )
24.
y =
2
e2x ln(4x)
4
x
¡ 2
¢
x ln(4x) + 1
2
y 0 = 4x e2x ln(4x) + e2x
2
=
4 e2x
x
2
28.
y =
y0 =
=
=
ex − e−x
ex + e−x
( ex + e−x )2 − ( ex − e−x )2
( ex + e−x )2
(2 e−x )(2 ex )
( ex + e−x )2
4
x
( e + e−x )2
34.
e−x
1
= ( ex + 1)−1
= x
−x
1+ e
e +1
y 0 = −( ex + 1)−2 ex
1
1
y 0 (0) = − 2 = −
2
4
y =
→
43.
S = 100000 e−0.5t
a)
S 0 = −50000 e−0.5t
b) The function is decreasing because S 0 < 0 for all t in addition to the argument in the
exponent being negative.
Math 1100-6
October 29, 2008
Page 4
§11.3
3.
8 = xy 2
→
0 = y 2 + 2xy
0 = 4+8
→
dy
dx
= −
dy
dx
dy
dx
Substitute in (2, 2)
1
2
9.
→
x2 + 4x + y 2 − 3y + 1
dy
dy
2x + 4 + 2y
−3
dx
dx
dy
(2y − 3)
dx
dy
dx
= 0
= 0
= −2(x + 2)
= −
2(x + 2)
2y − 3
16.
→
x2 − 3y 4
dy
→
2x − 12y 3
dx
3
2 dy
(−12y − 21y )
dx
dy
dx
= 2x5 + 7y 3 − 5
dy
= 10x4 + 21y 2
dx
= 10x4 − 2x
10x4 − 2x
12y 3 + 21y 2
2x(5x3 )
= − 2
3y (4y + 7)
= −
20.
(x + y)2
µ
¶
dy
→
2(x + y) 1 +
dx
£
¤ dy
⇒
2(x + y) − 15x4 y 2
dx
dy
dx
= 5x4 y 3
= 20x3 y 3 + 15x4 y 2
dy
dx
= 20x3 y 3 − 2(x + y)
=
20x3 y 3 − 2(x + y)
2(x + y) − 15x4 y 2
Math 1100-6
October 29, 2008
Page 5
32.
ln(x + y)
µ
¶
dy
1
1+
x+y
dx
·
¸
1
dy
− 2y
x+y
dx
dy
dx
→
= y2
dy
dx
1
= −
x+y
1
x+y
= −
·
x + y 1 − 2xy − 2y 2
1
= −
1 − 2xy − 2y 2
= 2y
38.
xy
x+
¶ e
µ
dy
exy
1+ y+x
dx
dy
x exy
dx
dy
→
dx
→
= 10
= 0
= −y exy − 1
= −
y exy + 1
x exy
47.
x2 + 4y 2 − 4x − 4 = 0
dy
→
2x + 8y
−4 = 0
dx
2−x
dy
=
⇒
dx
4y
a) Horizontal tangents occur when
(and y = 2).
dy
= 0, which means a horizontal tangent is at x = 2
dx
dy
→ ∞, which will occur when y → 0 (and
b) Vertical tangents occur when
dx
√
x = 2 ± 2 2).
57.
3
→
1
384 = (x + 1) 4 (y + 2) 3
1
1
3 1
2 dy
3
0 =
(x + 1)− 4 (y + 2) 3 + (x + 1) 4 (y + 2)− 3
4
3
dx
1
− 14
dy
9(x + 1) (y + 2) 3
= −
2
3
dx
4(x + 1) 4 (y + 2)− 3
= −
9(y + 1)
4(x + 1)
So, at x = 255 and at y = 214, we have
dy
dx
= −
9(215)
1935
=
≈ 1.8896
4(256)
1024