MTH 1125 Test #2 (1 pm) - Solutions
Fall 2004
Pat Rossi
Name
Instructions. Show CLEARLY how you arrive at your answers.
1. f (x) = 4x3 + 3x2 − 5x + 4; compute the slope of the line tangent to the graph of f (x)
at the point (2, 38) .
Slope, mtan = f 0 (x) = 12x2 + 6x − 5
At the point, (x, y) = (2, 38) , slope = f 0 (2) = 12 (2)2 + 6 (2) − 5 = 55
2.
d
dx
∙³
sec(x)
3x2 −5x+4
´10 ¸
=
Note: In the most general sense, this is a function raised to a power.
⎡
⎤
µ
¶9 µ ∙
¸¶
⎢µ sec (x) ¶10 ⎥
sec (x)
sec (x)
d
⎥
d ⎢
·
=
⎥ = 10
dx ⎢
3x2 − 5x + 4
dx 3x2 − 5x + 4
⎣ 3x2 − 5x + 4
⎦
|
{z
}
|
{z
} |
{z
}
(g(x))n
10
³
sec(x)
3x2 −5x+4
g 0 (x)
n(g(x))n−1
´9 µ sec (x) tan (x) (3x2 − 5x + 4) − (6x − 5) sec (x) ¶
·
(3x2 − 5x + 4)2
|
{z
}
quotient rule
3. x = sin (t) ; and t = 4y 2 − 2y + 6. Use the Liebniz form of the chain rule to compute
dx
.
dy
We Know:
dx
dt
dt
dy
= cos (t)
= 8y − 2
We Want:
dx
dy
=
i.e.,
dx
dy
dx dt
dt dy
= cos (t) (8y − 2)
= cos (4y 2 − 2y + 6) · (8y − 2)
¡
¢
4. f (x) = .|{z}
cos 2x3 + 5x2 . Compute f 0 (x) .
|
{z
}
outer
inner
¡
¢
¢ ¡
f 0 (x) = − sin 2x3 + 5x2 · 6x2 + 10x
|
{z
} |
{z
}
d eriv . o f o u ter
e va l. a t in n er
d eriv .
o f in n e r
=
-rewrite%
cos (4y 2 − 2y + 6) · (8y − 2)
5.
d
dx
[(2x3 − 3x + 5) (3x2 + 2)] =
Note: This is a product. We have to use the product rule.
⎡
⎤
¢¡ 2
¢⎥ ¡ 2
¢¡ 2
¢
¢
¡ 3
¡ 3
d ⎢
=
6x
−
3x
+
5
+
2
−
3
+
2
+
6x
−
3x
+
5
3x
3x
2x
2x
⎣
⎦
dx
|{z} |
|
| {z }| {z }
{z
}| {z }
{z
}
nd
1st
1st prime
2n d
2
2n d
prime
1st
6. x3 y 2 + 5x4 = 6y 3 − sin (y) ; compute y 0 .
(a)
1. Differentiate both sides w.r.t. x.
d
dx
[x3 y 2 + 5x4 ] =
d
dx
[6y 3 − sin (y)]
¡
¢
⇒ 3x2 · y 2 + 2yy 0 · x3 + 20x3 = 18y 2 y 0 − cos (y) y 0
|
{z
}
⇒
product rule
2. Solve algebraically for x.
1. Get the y 0 terms on the L.H. side. All other terms on the right.
⇒ 2yy 0 · x3 − 18y 2 y 0 + cos (y) y 0 = − (3x2 · y 2 + 20x3 )
2. Factor out y 0 .
⇒ y 0 (2y · x3 − 18y 2 + cos (y)) = − (3x2 · y 2 + 20x3 )
3. Divide by the “co-factor” of y 0 .
(3x2 ·y2 +20x3 )
⇒ y 0 = − (2y·x3 −18y2 +cos(y))
7. f (x) = 2x2 − 3x. Compute f 0 (x) , using the definition of derivative (i.e., using the
“limiting process.”)
f 0 (x) = lim∆x→0
f (x+∆x)−f (x)
∆x
= lim∆x→0
(2(x+∆x)2 −3(x+∆x))−(2x2 −3x)
lim∆x→0
(2(x2 +2x∆x+∆x2 )−3(x+∆x))−(2x2 −3x)
lim∆x→0
(4x∆x+2∆x2 −3∆x)
∆x
∆x
= lim∆x→0
∆x
= lim∆x→0
∆x(4x+2∆x−3)
∆x
2
=
(2x2 +4x∆x+2∆x2 −3x−3∆x)−(2x2 −3x)
∆x
= lim∆x→0 (4x + 2∆x − 3) = 4x − 3
=
Optional (Do this problem only if you want to be reimbursed for points lost on problem
#7 of Test #1)
Compute, using the properties of limits. Document each step!
∙
¸
(3x2 −5x+7)
limx→2 (5x3 +3x2 −2) =
limx→2 (3x2 − 5x + 7)
=
limx→2 (5x3 + 3x2 − 2)
|
{z
}
T h e lim it o f a q u otie nt e q u a ls
th e q u o tient o f th e lim its
(limx→2 3x2 ) − (limx→2 5x) + (limx→2 7)
=
(limx→2 5x3 ) + (limx→2 3x2 ) − (limx→2 2)
|
{z
}
T h e lim it o f a su m o r d iff e ren ce eq u a ls
th e su m o r d iff ere n c e of th e lim its
(limx→2 3x2 ) − (limx→2 5x) + 7
=
(limx→2 5x3 ) + (limx→2 3x2 ) − 2
|
{z
}
T h e lim it of a co n sta nt
is th e co n sta nt itself
(3 limx→2 x2 ) − (5 limx→2 x) + 7
=
(5 limx→2 x3 ) + (3 limx→2 x2 ) − 2
|
{z
}
T h e lim it o f a c on sta nt tim e s a fu n ctio n e q u a ls
th e co n sta nt tim es th e lim it o f th e fu n c tio n
3 (2)2 − 5 (2) + 7
=
5 (2)3 + 3 (2)2 − 2
|
{z
}
9
50
limx→c xn = cn
3