MTH131 Applied Calculus – Spring 2016
Lab 6 – Differentiation of Exponential and Logarithmic Functions
SOLUTIONS
1. (a) log2 4 = 2
(b) log4 2 = 1/2
1
(c) log2 = −2
4
(d) log8 4 = 2/3
(e) log6 3 + log6 12 = log6 [(3)(12)] = log6 36 = 2
(f) ln e2014 = 2014
2
(g) e2 ln 5 = eln 5 = eln 25 = 25
1
2
(h) ln(ex ) + 2 ln
= ln e + ln x2 − 2 ln x = 1 + 2 ln x − 2 ln x = 1
x
2. (a) f 0 (x) = (x2 )0 ln x + x2 (ln x)0 = 2x ln x + x2 (1/x) = 2x ln x + x
(b) f 0 (x) = (x)0 ln x2 + x(ln x2 )0 = ln x2 + x(1/x2 )(2x) = ln x2 + 2
(c) f 0 (x) = (x)0 (ln x)2 + x((ln x)2 )0 = (ln x)2 + x(2 ln x)(1/x) = (ln x)2 + 2 ln x
(d) f 0 (x) = ex
(e)
f 0 (x)
=
ex
2 −2x
(2x − 2)
+ xex − 2/x + 6(x2 + 1)2 x
3e3x + 7
(f) f 0 (x) = √
2 e3x + 7x
1
(g) ln 2 = ln x−2 = −2 ln x, so f 0 (x) = −2/x
x
1
(h)
= (ln x2 )−1 = (2 ln x)−1 , so f 0 (x) = −(2 ln x)−2 (1/x)
ln x2
3. (a) f 00 (x) = (2x)0 ln x + 2x(ln x)0 + (x)0 = 2 ln x + 2x(1/x) + 1 = 2 ln x + 3.
f 000 (x) = 2/x.
(b) f 00 (x) = (ln(x2 ) + 2)0 = (2 ln(x) + 2)0 = 2/x.
f 000 (x) = −2/x2 .
(c) f 00 (x) = ((ln x)2 )0 + (2 ln x)0 = 2(ln x)(1/x) + 2/x = 2/x(ln x + 1).
f 000 (x) = (2/x)0 (ln x + 1) + 2/x(ln x + 1)0 = −2/x2 (ln +1) + 2/x(1/x) = −(2 ln x)/x2 −
2/x2 + 2/x2 = −(2 ln x)/x2 .
4. (a) The formula for yearly compounding is P (1 + r)t . Using P = 10000, r = 0.1 and t = 10
we have 10000(1.1)10 = $25937.42.
(b) The formula for quarterly compounding is P (1 + r/4)4t . Using P = 10000, r = 0.1 and
t = 10 we have 10000(1.025)40 = $26850.64.
(c) The formula for monthly compounding is P (1 + r/12)12t . Using P = 10000, r = 0.1 and
t = 10 we have 10000(1.00833)120 = $27070.41
(d) The formula for continuous compounding is P ert . Using P = 10000, r = 0.1 and t = 10
we have 10000e1 = $27182.82.
5. The general formula for future value when compounding interest periodically is
F (t) = P (1 +
Since
d f (x)
dx a
r mt
)
m
= (ln a)af (x) f 0 (x), we have
"
0
F (t) =
d
r
P 1+
dx
m
r
= P ln 1 +
m
r
= P m ln 1 +
m
mt #
r
1+
m
mt
r
1+
m
d
(mt)
dx
mt
This allows us to answer the first three parts of this problem
(a) For all problems P = 10000 and r = 0.1. With yearly compounding m = 1 so the rate
of growth at time t = 10 is
0.1
0.1
1+
1
1
10
= 10000 ln(1.1)(1.1) = 2472.10
F 0 (10) = (10000)(1) ln 1 +
1(10)
(b) With quarterly compounding m = 4 so the rate of growth at time t = 10 is
0.1
0.1 4(10)
F (10) = (10000)(4) ln 1 +
1+
4
4
40
= 40000 ln(1.025)(1.025) = 2652.05
0
(c) With monthly compounding m = 12 so the rate of growth at time t = 10 is
0.1
0.1 12(10)
F (10) = (10000)(12) ln 1 +
1+
12
12
120
= 120000 ln(1.0083)(1.0083) = 2695.82
0
(d) The formula for continuous compounding is F (t) = P ert , so F 0 (t) = P rert . With the
values given, the rate of growth at time t = 10 is
F 0 (10) = (10000)(0.1)e0.1(10)
= 1000e1 = 2718.28
6. (a)
0.09 12t
= 2P
P 1+
12
1.007512t = 2
ln(1.007512t ) = ln 2
12t ln(1.0075) = ln 2
ln 2
0.6913
=
= 7.730
12 ln(1.0075)
12(0.007472)
t =
This equates to 7 years and 8.7 months, so it would take a total of 93 months
(b)
r
P 1+
12
1+
12(5)
= 2P
r
12
60
1+
= 2
r
= 21/60 = 1.012
12
r
= 0.012
12
r = 0.1394
or a rate of 13.94%.
2
7. (a) To maximize E(p) = 5000pe−0.005p we must set its derivative equal to 0.
2
2
E 0 (p) = 5000[(p)0 e−0.005p + p(e−0.005p )0 ] = 0
2
2
5000[e−0.005p + pe−0.005p (−0.005)(2p)] = 0
2
e−0.005p − 0.01p2 e−0.005p
2
= 0
1 − 0.01p
2
= 0
p
2
= 1/0.01 = 100
p = 10.
A price of $10 will maximize consumer expenditure.
2
(b) We are now maximizing E(p) = (5000)p2−0.005p .
2
2
E 0 (p) = 5000[(p)0 2−0.005p + p(2−0.005p )0 ] = 0
2
2
5000[2−0.005p + p(ln 2)2−0.005p (−0.005)(2p)] = 0
2
2
2−0.005p − 0.01(ln 2)p2 2−0.005p
= 0
2
= 0
2
= 1/(0.01 ln 2) = 144.27
1 − 0.01(ln 2)p
p
p = 12.01.
A price of $12.01 will maximize consumer expenditure.
8. Let f (t) = 11.274[t − 1.06(1 − e−t/1.06 )]
(a) f (5) = 11.274[5 − 1.06(1 − e−5/1.06 )] = 44.53 meters
(b) f 0 (t) = 11.274[1 − 1.06(−e−t/1.06 )(−1/1.06)] = 11.274(1 − e−t/1.06 ) meters/sec. So his
speed at t = 5 seconds was f 0 (5) = 11.274(1 − e−5/1.06 ) = 11.173 meters/sec
(c) To answer this question, we first have to determine the time at which his speed was 9
meters/sec.
11.274(1 − e−t/1.06 ) = 9
1 − e−t/1.06 = 0.798
e−t/1.06 = 0.202
−t/1.06 = ln(0.202)
t = −1.06 ln(0.202) = 1.70 secs
Plugging 1.70 into the original distance formula f (t) gives
f (1.7) = 11.274[1.70 − 1.06(1 − e−1.70/1.06 )]
= 9.6 meters
9. (a) Using the demand function D(p) = 49e−0.004p , we have
D0 (p) = 49e−0.004p (−0.004)
= −0.196e−0.004p
p(−0.196e−0.004p )
El(p) = −
49e−0.004p
= 0.004p
When p = 200, El(200) = 0.8 > 1, indicating that the demand is inelastic and the
manufacturer can raise his price to increase revenue. The price that leads to unitelasticity is
0.004p = 1
p =
1
= $250
0.004
(b) Using the demand function D(p) = (2401 − (1/2)p3/2 )1/2 , we have
1
1
D (p) =
(2401 − (1/2)p3/2 )−1/2 −
2
2
1/2
3p
= −
8(2401 − (1/2)p3/2 )1/2
0
El(p) = −
=
1/2
3p
p − 8(2401−(1/2)p
3/2 )1/2
(2401 − (1/2)p3/2 )1/2
3p3/2
8(2401 − (1/2)p3/2 )
3 1/2
p
2
When p = 200, El(200) = 1.075 > 1, indicating that the demand is (slightly) elastic
and the manufacturer can lower his price to increase revenue. The price that leads to
unit-elasticity is
3p3/2
8(2401 − (1/2)p3/2 )
= 1
3p3/2 = 8(2401 − (1/2)p3/2 )
= 19208 − 4p3/2
7p3/2 = 19208
p3/2 = 2744
p = 27442/3
p = $196