MTH 1125 Practice Test #2 - Solutions
Fall 1987
Pat Rossi
Name
1. Compute: limx→2
(a) limx→2
(2x3 +4x)(x2 −3)
x3 +2x2 +x+1
(2x3 +4x)(x2 −3)
x3 +2x2 +x+1
=
, using the properties of limits. Document each step.
limx→2 [(2x3 +4x)(x2 −3)]
limx→2 [x3 +2x2 +x+1]
The limit of a quotient equals
the quotient of the limits
b
=
[limx→2 2x3 +limx→2 4x][limx→2 x2 −limx→ 3]
limx→2 x3 +limx→2 2x2 +limx→2 x+limx→2 1
The limit of a sum or difference
is the sum or difference of the limits
2(limx→2 x)3 +4 limx→2 x limx→2 x2 −limx→2 3
(limx→2 x)3 +2(limx→2 x)2 +limx→2 x+limx→2 1
limx→a xn = an
b
[
3
=
[2(2)
][(
b
+4(2)][(2)2 −3]
)
c
]
=
=
c
=
(2)3 +2(2)2 +(2)+1
c
=
[limx→2 (2x3 +4x)][limx→2 (x2 −3)]
limx→2 [x3 +2x2 +x+1]
limit of a product
bequalsThe
the product of the limits c
[2 limx→2 x3 +4 limx→2 x][limx→2 x2 −limx→2 3]
limx→2 x3 +2 limx→2 x2 +limx→2 x+limx→2 1
[
=
The limit of a constant times a function equals
the constant times the limit of the function
2(2)3 +4(2) (2)2 −limx→2 3
(2)3 +2(2)2 +(2)+limx→2 1
blimx→a x = ac
b
][
]
c
24
19
limit of a constant is
bThethe
c
constant itself
2. f (x) = 2x2 − 2. Compute f 0 (x) using the definition of derivative.
[2(x+4x)2 −2]−[2x2 −2]
(x)
(a) f 0 (x) = lim4x→0 f (x+4x)−f
=
lim
4x→0
4x
4x
2
2 −2−2x2 +2
[2(x2 +2x4x+4x2 )−2]−[2x2 −2]
= lim4x→0
= lim4x→0 2x +4x4x+24x
4x
4x
2
= lim4x→0 4x4x+24x
= lim4x→0 4x(4x+24x)
4x
4x
= lim4x→0 (4x + 24x) = 4x + 2 (0) = 4x
Now plug in
3. f (x) = 3x4 − 2x2 + 6x + 8. Compute f 0 (x) using the rules and properties of differentiation. Document each step.
d
d
d
d
[3x4 − 2x2 + 6x + 8] = dx
[3x4 ] − dx
[2x2 ] + dx
[6x] + dx
[8]
The derivative of a sum or difference is the
b is the sum or difference of the derivatives c
d
d
d
d
d
= 3 · dx
[x4 ] − 2 · dx
[x2 ] + 6 dx
[x] + dx
[8] = 3 [4x3 ] − 2 [2x] + 6 [1] + dx
[8]
The derivative of a constant times a function equals
Power
b the constant times the derivative of the function c
b rule c
= 3 [4x3 ] − 2 [2x] + 6 [1] + 0 = 12x3 − 4x + 6
derivative of
bofThea constant
is 0c
(a) f 0 (x) =
d
dx
4. Compute the slope of the line tangent to the graph of f (x) = 6x2 − 3x + 2 at the point
(2, 20) .
(a) f (x) = 6x2 − 3x + 2
mtan = f 0 (x) = 12x − 3
(mtan at x = 2) is f 0 (2) = 12 (2) − 3 = 21
5. f (x) = (x2 − 3) (2x3 + 2x) .Compute f 0 (x) using the Product rule
¡
¢¡
¢
(a) f (x) = x2 − 3 2x3 + 2x
| {z }| {z }
nd
1st
¢¡
¢
¡ 3 2 ¢ ¡ 2
0
f (x) = (2x) 2x + 2x + 6x + 2 x2 − 3
|{z} | {z } | {z }| {z }
1st
2nd
bofderiv
bofderiv
1st c
2nd c
6. f (x) = (x2 − x) (2x3 − 2x2 ) (3x4 − 3x3 ) . Compute f 0 (x) using a generalized form of
the product rule.
¡
¢¡
¢¡
¢
(a) f (x) = x2 − x 2x3 − 2x2 3x4 − 3x3
| {z }|
{z
}|
{z
}
1st
2nd
3rd
¢¡
¢ ¡
¡
¢¡
¢ ¡
¢
f 0 (x) = (2x − 1) 2x3 − 2x2 3x4 − 3x3 + 6x2 − 4x x2 − x 3x4 − 3x3 +
| {z }|
{z
}|
{z
} | {z }| {z }|
{z
}
deriv of 1st
2nd
3rd
1st
deriv of 3rd
7. f (x) =
x4 −2x2
x2 −2x+3
(a) f (x) =
f (x) =
3rd
2nd
; compute f 0 (x)
x4 −2x2←−top
x2 −2x−3←−bottom
↓top prime
0
1st
deriv of 2nd
¡
¢¡
¢¡
¢
12x3 − 9x2 x2 − x 2x3 − 2x2
|
{z
}| {z }|
{z
}
↓bottom
↓bottom prime
(4x3 −4x)(x2 −2x+3) −
(2x−2)
(x2 −2x+3)2
↓top
(x4 −2x2 )
↑
bottom squared
8. f (x) =
√1
x+1
(a) f (x) =
; compute f 0 (x) using the definition of derivative.
√1
x+1
f 0 (x) = lim4x→0
lim4x→0
f (x+4x)−f (x)
4x
1
1
− √x+1
(x+4x)+1
√
4x
=
= lim4x→0
¶2 ³
´2
1
1
√
− √x+1
(x+4x)+1
µ
¶
lim4x→0
1
1
4x √
+ √x+1
µ
µ
1
1
− √x+1
(x+4x)+1
√
= lim4x→0
1
· x+1 − 1 · x+4x+1 )
( x+4x+1
´
³ x+1 x+1 x+4x+1
1
1
√
+ √x+1
x+4x+1
combine terms on top,by
getting a common deminator
−4x
³(x+4x+1)(x+1) ´
1
1
4x √x+4x+1
+ √x+1
b
= lim4x→0
= lim4x→0
√
x+1
2
=−
4x
(
)
−1
(
)
³ (x+4x+1)(x+1) ´
√
1
1
+ √x+1
x+4x+1
·
³
µ
√
1
1
+ √x+1
x+4x+1
´
1
1
+ √x+1
(x+4x)+1
√
¶
=
bmultiply top and bottom by radical conjugatec
(x+4x)+1
lim4x→0
4x
¶
( 1 − 1 ) ´
³ x+4x+1 x+1
1
1
4x √x+4x+1
+ √x+1
(³(x+1)−(x+4x+1)
(x+4x+1)(x+1) )
´
= lim4x→0
c
= lim4x→0
4x
1
3
2
√
1
1
+ √x+1
x+4x+1
−4x
(³(x+4x+1)(x+1)
)
4x
Now plug in! =
2(x+1) 2
=
1
1
√
+ √x+1
x+4x=1
´
−1
(
)
³ (x+0+1)(x+1) ´
1
√ 1
+ √x+1
x+0+1
=
³
´
−1
2
³(x+1) ´
2
√
x+1
=
−1
(x+1)2
·
9.
d
dx
[(4x5 − 6x2 + 3x − 7)20 ] =
(a)
10.
d
dx
d
dx
∙³
(a)
¸
∙
20
¢
¡
¢19 ¡
¢
¡ 5
2
4x − 6x + 3x − 7 = 20 4x5 − 6x2 + 3x − 7 · 20x4 − 12x + 3
|
{z
}
|
{z
} |
{z
}
0
n−1
[g(x)n ]
(x)
g
n [g (x)]
| {z }
| {z }
deriv of
Power Rule
b
inner functionc
b as usual c
3x4 +2x
4x3 +3x2
d
dx
∙³
´10 ¸
3x4 +2x
4x3 +3x2
=
´10 ¸
observe: For the purpose of classification, this is a function raised
to a power. Use the Chain rule:
¶9
¸
∙
µ 4
d 3x4 + 2x
3x + 2x
·
Now , and only now that our problem has
= 10
4x3 + 3x2
dx 4x3 + 3x2
|
{z
} |
{z
}
power rule as usual
deriv of (inner)function
h 4
i
d
3x +2x
, do we consider using the Quotient
been reduced to that of computing dx
4x3 +3x2
Rule:
³ 4
´9 µ (12x3 + 2) (4x3 + 3x2 ) − (12x2 + 6x) (3x4 + 2x) ¶
3x +2x
= 10 4x3 +3x2 ·
(4x3 + 3x2 )2
|
{z
}
By the Quotient Rule
11. y = u5 ; u =
(a)
dy
dx
dy
du
du
dx
=
dy
1
;compute dx
3x−2
dy
du
·
du
dx
using the Liebniz form of The Chain Rule.
so we need to find
dy
du
and
du
dx
= 5u4
=
0·(3x−2)−3·(1)
(3x−2)2
=
−3
(3x−2)2
³
´
dy
dy du
3
= du
· dx = 5u4 · − (3x−2)
2
dx
Now, express derivative solely in terms of x
dy
dx
´
´
³
¡ 1 ¢4 ³
¡ 1 ¢
3
3
15
= 5u · − (3x−2)2 = 5 3x−2 = 5 3x−2 · − (3x−2)2 = − (3x−2)
6
i.e.
4
dy
dx
15
= − (3x−2)
6
⎡
⎤
h
i
¢5 ¡ 10 ¢⎥
¡
5
10
dy
d ⎢
12. dx
(x4 + 6x) (3x3 + 4x)
= dx
⎣ x4 + 6x 3x3 + 4x ⎦ Observe that in the most
| {z }| {z }
1st
2nd
general
the function
to be differentiated
µ sense,
¶
¶ Use the Product rule.
µ h is a Product.
¢5 i ¡ 3
¢10
¢10 i ¡ 4
¢5
d h¡ 4
d ¡ 3
=
x + 6x
3x + 4x
3x + 4x +
x + 6x
dx
dx
|
{z
}
| {z }
{z
}
{z
}
|
|
1st
2nd
deriv of 1st
|
deriv of 2nd
{z
By Product Rule
3
}
Now, and only now that our problem has been reduced to that of computing the
derivative of a function to a power, do we use the Chain Rule:
h ¡
h ¡
¢4 ¡
¢i
¢9 ¡
¢i
10
5
= 5 x4 + 6x · 4x3 + 6 (3x3 + 4x) + 10 3x3 + 4x · 9x2 + 4 (x4 + 6x)
{z
}
{z
}
|
|
By Chain Rule
13. Compute:
d
dx
⎡
By Chain Rule
[sin (3x2 − 6)]
⎤
¡
¢
¢
¡
2
d ⎣
⎦ = cos 3x2 − 6 · (6x)
3x
−
6
(a) dx
sin
|{z}| {z }
|
{z
} |{z}
outer
of
of outer
inner
bderiv.
bderiv
c
inner c
eval. at inner
14. Compute:
d
dx
[tan3 (x)]
(a) Rewrite:
d
dx
[tan3 (x)] =
¤
d £
(tan (x))3 = 3 (tan (x))2 · sec2 (x) = 3 tan2 (x) sec2 (x)
| {z } | {z }
|dx
{z
}
n(g(x))n−1
d
[(g(x))n ]
dx
0
0
g (x)
15. Suppose that E (x) has the property that E (x) = E (x) . Compute
(a)
d
dx
⎡
⎤
⎣ E (tan (x))⎦ = E (tan (x)) · sec2 (x)
|{z}| {z }
| {z } | {z }
outer
of outer
of
inner
bderiv
bderiv
eval. at inner c
inner c
4
d
dx
[E (tan (x))] .