MTH 1125 - Test 2 (10 am class) - Solutions
Fall 2011
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. Compute:
d
dx
d
dx
√
[6x5 + 5x4 + 15x2 + 20x + 30 x + 50] =
i
h
√
1
[6x5 + 5x4 + 15x2 + 20x + 30 x + 50] = 6 [5x4 ] + 5 [4x3 ] + 15 [2x] + 20 + 30 12 x− 2
1
= 30x4 + 20x3 + 30x + 20 + 15x− 2
i.e.,
d
dx
√
1
[6x5 + 5x4 + 15x2 + 20x + 30 x + 50] = 30x4 + 20x3 + 30x + 20 + 15x− 2
d
2. Compute: dx
[(x3 + x2 ) tan (x)] =
⎡
⎤
¢
¡
¢
¢
¡
¡
3
2
d ⎣
⎦ = 3x2 + 2x · tan (x) + sec2 (x) · x3 + x2
+
x
tan
(x)
x
dx
| {z } | {z } | {z } | {z }
| {z }| {z }
2nd
1st
d
dx
2nd
1st prime
2n d prime
1st
[(x3 + x2 ) tan (x)] = (3x2 + 2x) · tan (x) + sec2 (x) · (x3 + x2 )
3. Compute:
⎡
d
dx
h
cos(x)
3x3 +2x
i
=
⎤
b o tto m p rim e
b o tto m
to p p rim e
to p
to p
z }| { ¡z }| ¢{ z¡ }| ¢{ z }| {
z }| {
3
2
(− sin (x)) · 3x + 2x − 9x + 2 · cos (x)
d ⎢ cos (x) ⎥
=
¢2
¡ 3
dx ⎣ 3x3 + 2x ⎦
3x + 2x
| {z }
|
{z
}
B o tto m
b o tto m
sq u a red
i.e.,
d
dx
h
cos(x)
3x3 +2x
i
=−
sin(x)·(3x3 +2x) + (9x2 +2)·cos(x)
(3x3 +2x)2
4. Compute:
d
dx
d
dx
h
i
10
4
3
(6x + 8x + 24x + 2) =
This is a function raised to a power.
h
i
¢9 ¡
¢
¡
10
(6x4 + 8x3 + 24x + 2) = 10 6x4 + 8x3 + 24x + 2 · 24x3 + 24x2 + 24
|
{z
} |
{z
}
p ow er ru le
a s u su a l
i.e.,
d
dx
d eriva tive
o f in n e r
h
i
10
9
(6x4 + 8x3 + 24x + 2) = 10 (6x4 + 8x3 + 24x + 2) (24x3 + 24x2 + 24)
5. Given that f (x) = 3x2 − 4x + 2, give the equation for the line tangent to the graph of
f (x) at the point (1, 1) .
We need two things:
(a)
1. A point on the line (We have that: (x1 , y1 ) = (1, 1))
2. The slope of the line (This is f 0 (x1 ))
f 0 (x) = 6x − 4
The slope at the point (x1 , y1 ) = (1, 1) is f 0 (1) = 6 (1) − 4 = 2
We will use the Point-Slope equation of a line: y − y1 = m (x − x1 )
Thus, the equation of the line tangent to the graph of f (x) is:
y − 1 = 2 (x − 1)
The equation of the line tangent is y − 1 = 2 (x − 1)
2
6. Given that x = sec (t) and that t = sin (u) + u2 ; compute
of the Chain Rule.
dx
du
using the Liebniz form
We know:
dx
dt
= sec (t) tan (t)
dt
du
= cos (u) + 2u
We want:
dx
du
By the Liebniz form of the Chain Rule:
dx
du
=
dx dt
dt du
¡
¢
¡
¢
= sec (t) tan (t) (cos (u) + 2u) = sec sin (u) + u2 tan sin (u) + u2 (cos (u) + 2u)
|
{z
}
e x p ress so lely in te rm s o f
in d ep en d ent va ria b le u
i.e.
dx
du
= sec (sin (u) + u2 ) tan (sin (u) + u2 ) (cos (u) + 2u)
7. Compute:
d
dx
[sec (5x3 + 25x)] =
Outer: = sec
Deriv. of outer
d
dx
⎡
⎢
⎢
⎢
⎢
⎣
i.e.,
= sec
3
sec (5x + 25x)
%
↑
outer
inner
d
dx
¡
¡
¢
¢
tan
⎤
¡
¢
⎥
⎥
¢
¡
¢ ¡
¢
¡
⎥ = sec 5x3 + 25x tan 5x3 + 25x · 15x2 + 25
⎥ |
{z
} |
{z
}
⎦
D e riv o f o u ter,
d e riv . o f
e va l. a t in n er
[sec (5x3 + 25x)] = sec (5x3 + 25x) tan (5x3 + 25x) · (15x2 + 25)
3
in n e r
d
dx
8. Compute:
power.
d
dx
∙³
2x2 +4
6x3 +9x2
∙³
2x2 +4
6x3 +9x2
´20 ¸
i.e.,
∙³
In the broadest sense, this is a function raised to a
=
µ
2x2 + 4
= 20
6x3 + 9x2
|
{z
p ow er ru le
a s u su a l
= 20
d
dx
´20 ¸
2x2 +4
6x3 +9x2
´20 ¸
³
2x2 +4
6x3 +9x2
= 20
¶19 µ ∙
¸¶
2x2 + 4
d
·
dx 6x3 + 9x2
} |
{z
}
d e riva tive
o f in n er
´19 (4x) (6x3 + 9x2 ) − (18x2 + 18x) (2x2 + 4)
·
(6x3 + 9x2 )2
|
{z
}
Q u o tient
R u le
³
2x2 +4
6x3 +9x2
´19
·
(4x)(6x3 +9x2 )−(18x2 +18x)(2x2 +4)
(6x3 +9x2 )2
d
9. Compute: dx
[tan5 (4x3 + 12x)] =
Re-write!
h
i
5
d
3
(tan (4x + 12x)) =
This is a function, raised to a power
dx
d
dx
µ
¶
h
i
¡ 3
¡
¡ 3
¢¢4
¢¤
d £
5
3
tan 4x + 12x
(tan (4x + 12x)) = 5 tan 4x + 12x
·
dx
|
{z
}
{z
}
|
p ow er ru le
a s u su a l
d e riva tive
o f in n er
¢ ¡
¢
¡
4
= 5 (tan (4x3 + 12x)) · sec2 4x3 + 12x · 24x2 + 12
|
{z
}
C h ain
R u le
i.e.,
d
dx
4
[tan5 (4x3 + 12x)] = 5 (tan (4x3 + 12x)) · sec2 (4x3 + 12x) · (24x2 + 12)
4
10. Given that f (x) = 3x2 + 2x + 5,compute f 0 (x) using the definition of derivative.
f 0 (x) = lim∆x→0
f (x+∆x)−f (x)
∆x
= lim∆x→0
[3(x+∆x)2 +2(x+∆x)+5]−[3x2 +2x+5]
∆x
= lim∆x→0
[3(x2 +2x∆x+∆x2 )+2(x+∆x)+5]−[3x2 +2x+5]
= lim∆x→0
[3x2 +6x∆x+3∆x2 +2x+2∆x+5]−[3x2 +2x+5]
= lim∆x→0
∆x(6x+3∆x+2)
∆x
∆x
∆x
= lim∆x→0
6x∆x+3∆x2 +2∆x
∆x
= lim∆x→0 6x + 3∆x + 2 = 6x + 2
i.e., f 0 (x) = 6x + 2
5