Jim Lambers
MAT 167
Fall Semester 2011-12
Practice Exam 2 Solution
1. Given y = f (x), where
f (x) = −12x2 + 48x + 24,
compute the instantaneous rate of change of y with respect to x, or dy/dx, at the point x = 3.
Solution We first compute dy/dx, using the Power Rule on each term:
dy
dx
d
[−12x2 + 48x + 24]
dx
d
d
d
=
[−12x2 ] +
[48x] +
[24]
dx
dx
dx
d
d
d
= −12 [x2 ] + 48 [x] + 24 [1]
dx
dx
dx
= −12(2x) + 48(1) + 24(0)
=
= −24x + 48.
The instantaneous rate of change of y with respect to x at x = 3 is given by the value of
dy/dx at x = 3, which is
f 0 (3) = −24(3) + 48 = −72 + 48 = −24.
2. Given
f (x) = (x − 1)3 ,
compute the function f 0 (x) using the definition of the derivative. Note: you may not use
differentiation rules.
Solution Using the definition of the derivative,
f (x + h) − f (x)
,
h→0
h
f 0 (x) = lim
we have
[(x + h) − 1]3 − (x − 1)3
h→0
h
[(x − 1) + h]3 − (x − 1)3
= lim
h→0
h
3
(x − 1) + 3h(x − 1)2 + 3h2 (x − 1) + h3 − (x − 1)3
= lim
h→0
h
f 0 (x) =
lim
1
3h(x − 1)2 + 3h2 (x − 1) + h3
h→0
h
= lim 3(x − 1)2 + 3h(x − 1) + h2
=
lim
h→0
= 3(x − 1)2 .
3. Differentiate the function
f (x) =
(x + 1)2 (x + 2)2
.
x3 − 3x2 + 3x − 1
Solution Using the Quotient Rule, as well as the Product Rule, we obtain
(x + 1)2 (x + 2)2
d
f 0 (x) =
dx x3 − 3x2 + 3x − 1
=
=
=
d
d
(x3 − 3x2 + 3x − 1) dx
[(x + 1)2 (x + 2)2 ] − (x + 1)2 (x + 2)2 dx
[x3 − 3x2 + 3x − 1]
(x3 − 3x2 + 3x − 1)2
d
d
(x3 − 3x2 + 3x − 1) (x + 2)2 dx
[(x + 1)2 ] + (x + 1)2 dx
[(x + 2)2 ]
−
(x3 − 3x2 + 3x − 1)2
(x + 1)2 (x + 2)2 (3x2 − 6x + 3)
(x3 − 3x2 + 3x − 1)2
(x3 − 3x2 + 3x − 1) (x + 2)2 [2(x + 1)] + (x + 1)2 [2(x + 2)]
−
(x3 − 3x2 + 3x − 1)2
(x + 1)2 (x + 2)2 (3x2 − 6x + 3)
.
(x3 − 3x2 + 3x − 1)2
4. Differentiate the function
cos x csc x
.
sec x
Solution Using the Quotient Rule, the Product Rule, and the rules for differentiating trigonometric functions, we obtain
d h cos x csc x i
f 0 (x) =
dx
sec x
d
d
[sec x]
sec x dx [cos x csc x] − cos x csc x dx
=
2x
sec
d
d
sec x csc x dx
[cos x] + cos x dx
[csc x] − cos x csc x(sec x tan x)
=
sec2 x
sec x [csc x(− sin x) + cos x(− csc x cot x)] − cos x csc x sec x tan x
=
.
sec2 x
sec x [− csc x sin x − cos x csc x cot x] − cos x csc x sec x tan x
=
.
sec2 x
f (x) =
2
5. Differentiate the function
f (x) = cos(sec3 (x4 )).
Solution Using the General Power Rule and the Chain Rule, we obtain
d cos(sec3 (x4 ))
dx
d
= − sin(sec3 (x4 )) [sec3 (x4 )]
dx
d
3 4
= − sin(sec (x ))3 sec2 (x4 ) [sec(x4 )]
dx
f 0 (x) =
d 4
[x ]
dx
= − sin(sec3 (x4 ))3 sec2 (x4 ) sec(x4 ) tan(x4 )4x3 .
= − sin(sec3 (x4 ))3 sec2 (x4 ) sec(x4 ) tan(x4 )
6. Given
y 2 (y 2 − 7) = x2 (x2 + 5),
compute the equation of the tangent line to the curve described by this equation at the point
(2, 3).
Solution Differentiating the given equation implicitly with respect to x, using the Chain
Rule and the Product Rule, yields
(y 2 − 7)(2y)
dy
dy
+ y 2 (2y)
= (x2 + 5)(2x) + x2 (2x).
dx
dx
We then substitute x = 2 and y = 3 to obtain
dy
dy
+ 32 (2(3))
dx
dx
dy
dy
2(6)
+ 9(6)
dx
dx
dy
dy
12
+ 54
dx
dx
dy
66
dx
dy
dx
(32 − 7)(2(3))
= (22 + 5)(2 · 2) + 22 (2 · 2)
= 9(4) + 4(4)
= 36 + 16
= 52
=
26
.
33
We then use the point-slope form,
y − y0 = m(x − x0 ),
3
with m = 26/33, x0 = 2, and y0 = 3 to obtain
y−3=
or
y=
26
(x − 2),
33
47
26
x+ .
33
33
7. Recall the equation from Problem 6,
y 2 (y 2 − 7) = x2 (x2 + 5).
Compute y 00 .
Solution From Problem 6, we have
(y 2 − 7)(2y)
dy
dy
+ y 2 (2y)
= (x2 + 5)(2x) + x2 (2x),
dx
dx
which yields
(x2 + 5)(2x) + x2 (2x)
4x3 + 10x
dy
= 2
=
.
dx
(y − 7)(2y) + y 2 (2y)
4y 3 − 14y
To obtain y 00 = d2 y/dx2 , we differentiate dy/dx using the Quotient Rule and Chain Rule to
obtain
d dy
00
y =
dx dx
d 4x3 + 10x
=
dx 4y 3 − 14y
=
d
d
(4y 3 − 14y) dx
[4x3 + 10x] − (4x3 + 10x) dx
[4y 3 − 14y]
(4y 3 − 14y)2
=
dy
(4y 3 − 14y)(12x2 + 10) − (4x3 + 10x)(12y 2 − 14) dx
(4y 3 − 14y)2
3
=
+10x
(4y 3 − 14y)(12x2 + 10) − (4x3 + 10x)(12y 2 − 14) 4x
4y 3 −14y
(4y 3 − 14y)2
,
where in the last step we have substituted the formula for dy/dx obtained previously.
8. Given
f (x) = x cos2 (2x),
compute f 00 (π).
4
Solution Using the Product Rule, we obtain the first derivative,
d
[x cos2 (2x)]
dx
d
d
= cos2 (2x) [x] + x [cos2 (2x)]
dx
dx
d
2
= cos (2x)(1) + x 2 cos(2x) [cos(2x)]
dx
f 0 (x) =
= cos2 (2x) − 2x cos(2x) sin(2x)(2)
= cos(2x)[cos(2x) − 4x sin(2x)].
To compute f 00 (x), we use the Product Rule on f 0 (x) and obtain
f 00 (x) =
=
=
=
=
=
d
[cos(2x)[cos(2x) − 4x sin(2x)]]
dx
d
d
[cos(2x) − 4x sin(2x)] [cos(2x)] + cos(2x) [cos(2x) − 4x sin(2x)]
dx
dx
d
d
−[cos(2x) − 4x sin(2x)] sin(2x)(2) + cos(2x)
[cos(2x)] −
[4x sin(2x)]
dx
dx
d
−2[cos(2x) − 4x sin(2x)] sin(2x) + cos(2x) −2 sin(2x) − 4 [x sin(2x)]
dx
−2[cos(2x) − 4x sin(2x)] sin(2x) +
d
d
cos(2x) −2 sin(2x) − 4 sin(2x) [x] + x [sin(2x)]
dx
dx
−2[cos(2x) − 4x sin(2x)] sin(2x) +
cos(2x) [−2 sin(2x) − 4 [sin(2x)(1) + x(cos(2x))(2)]]
= −2[cos(2x) − 4x sin(2x)] sin(2x) + cos(2x) [−2 sin(2x) − 4 [sin(2x) + 2x cos(2x)]]
= −2[cos(2x) − 4x sin(2x)] sin(2x) − cos(2x) [2 sin(2x) + 4 [sin(2x) + 2x cos(2x)]] .
Since sin(2π) = 0 and cos(2π) = 1, it follows that
f 00 (2) = −2[cos(2π) − 4π sin(2π)] sin(2π) − cos(2π) [2 sin(2π) + 4 [sin(2π) + 2π cos(2π)]]
= −2[1 − 4π(0)](0) − (1) [2(0) + 4 [0 + 2π(1)]]
= −1 [4 [2π]]
= −8π.
9. Suppose that the radius r of a cylinder is increasing at a constant rate of 2 cm/s, while its
height h is decreasing at a constant rate of 1 cm/s. Determine the rate at which the volume
V of the cylinder is changing when r = 5 cm and h = 100 cm.
5
Solution From the formula for the volume of a cylinder,
V = πr2 h,
we obtain, through differentiation with respect to t and the Product Rule,
dr
dh
dV
= π2rh + πr2 .
dt
dt
dt
Substituting r = 5, h = 100, dr/dt = 2 and dh/dt = −1, we obtain
dV
= π2(5)(100)(2) + π(52 )(−1) = 2000π − 25π = 1975π cm3 /s.
dt
10. Compute
1/x
1
lim
.
−
2
x→0
Solution First, we note that
1/x
1/x
1
= 2−1
= 2−1/x .
2
Second, we note that because x < 0, we have
lim −
x→0−
1
= ∞.
x
Because
lim 2t = ∞,
t→∞
we conclude that
1/x
1
lim
= lim 2−1/x = lim 2t = ∞,
−
t→∞
2
x→0
x→0−
where t = −1/x.
6