Chapter 11 – 3 Derivatives of Products and Quotients
Example 1. Find the derivative of 𝑓(𝑥) = 𝑥 2 𝑒 𝑥
Find the derivative of F(x)G(x):
𝐹(𝑥 + ℎ)𝐺(𝑥 + ℎ) − 𝐹(𝑥)𝐺(𝑥)
ℎ→0
ℎ
lim
= lim
𝐹(𝑥 + ℎ)𝐺(𝑥 + ℎ) − 𝐹(𝑥 + ℎ)𝐺(𝑥) + 𝐹(𝑥 + ℎ)𝐺(𝑥) − 𝐹(𝑥)𝐺(𝑥)
ℎ
ℎ→0
= lim
𝐹(𝑥 + ℎ)[𝐺(𝑥 + ℎ) − 𝐺(𝑥)] + [𝐹(𝑥 + ℎ) − 𝐹(𝑥)]𝐺(𝑥)
ℎ
ℎ→0
= lim
𝐹(𝑥 + ℎ)[𝐺(𝑥 + ℎ) − 𝐺(𝑥)]
ℎ
ℎ→0
= lim 𝐹(𝑥 + ℎ) lim
ℎ→0
+ lim
[𝐹(𝑥 + ℎ) − 𝐹(𝑥)]𝐺(𝑥)
ℎ
ℎ→0
𝐺(𝑥 + ℎ) − 𝐺(𝑥)
ℎ
ℎ→0
+ (lim
𝐹(𝑥 + ℎ) − 𝐹(𝑥)
ℎ→0
ℎ
) 𝐺(𝑥)
= 𝐹(𝑥)𝐺′ (𝑥) + 𝐹′ (𝑥)𝐺(𝑥)
In our problem, F(x) = x2 and G(x)=ex:
′
′
𝑓 ′ (𝑥) = 𝐹(𝑥)𝐺 (𝑥) + 𝐹 (𝑥)𝐺(𝑥) = 𝑥2 𝑒𝑥 + 2𝑥𝑒𝑥 = (𝑥2 + 2𝑥)𝑒𝑥
Example 2. Find the derivative of 𝑓(𝑥) = (𝑥 2 − 2)(2𝑥 + 3) two
different ways.
A) Write f(x) as 𝑓(𝑥) = 2𝑥 3 + 3𝑥 2 − 4𝑥 − 6.
𝑓 ′ (𝑥) = 6𝑥 2 + 6𝑥 − 4
B) Apply the product rule where 𝐹(𝑥) = 𝑥 2 − 2 and 𝐺(𝑥) = 2𝑥 + 3
𝐹(𝑥)𝐺 ′ (𝑥) + 𝐹 ′ (𝑥)𝐺(𝑥) = (𝑥 2 − 2)2 + 2𝑥(2𝑥 + 3)
= 2𝑥 2 − 4 + 4𝑥 2 + 6𝑥 = 6𝑥 2 + 6𝑥 − 4
Example 3. Find the derivative of 𝑓(𝑥) = 5𝑥 ln 𝑥
30
𝐹(𝑥) = 5𝑥 and 𝐺(𝑥) = ln 𝑥
1
𝑓 ′ (𝑥) = 5𝑥 + 5 ln 𝑥 = 5 + 5 ln 𝑥
𝑥
25
20
15
5xlnx
10
5
5+5lnx
0
-5 0
-10
-15
1
2
3
4
Example 4. Find the derivative of 𝑓(𝑥) =
𝑒𝑥
𝑥2
𝐹(𝑥) = 𝑒 𝑥 and 𝐺(𝑥) = 𝑥 −2
𝑓 ′ (𝑥) = 𝑒 𝑥 (−2𝑥 −3 ) + 𝑒 𝑥 𝑥 −2
−2𝑒 𝑥 𝑒 𝑥 −2𝑒𝑥 𝑥𝑒𝑥 𝑥𝑒𝑥 − 2𝑒𝑥 (𝑥 − 2)𝑒𝑥
=
+ 2=
+ 3 =
=
𝑥3
𝑥
𝑥3
𝑥
𝑥3
𝑥3
If 𝑓(𝑥) =
𝑓 ′ (𝑥)
𝑁(𝑥)
𝐷(𝑥)
then
𝑁(𝑥 + ℎ)𝐷(𝑥) 𝑁(𝑥)𝐷(𝑥 + ℎ)
𝑁(𝑥 + ℎ) 𝑁(𝑥)
−
−
𝐷(𝑥 + ℎ)𝐷(𝑥) 𝐷(𝑥)𝐷(𝑥 + ℎ)
𝐷(𝑥 + ℎ) 𝐷(𝑥)
= lim
= lim
ℎ→0
ℎ→0
ℎ
ℎ
1
𝑁(𝑥 + ℎ)𝐷(𝑥) − 𝑁(𝑥)𝐷(𝑥 + ℎ)
= lim
lim
ℎ→0 𝐷 (𝑥 + ℎ )𝐷(𝑥) ℎ→0
ℎ
=
1
𝑁(𝑥 + ℎ)𝐷(𝑥) − 𝑁(𝑥)𝐷(𝑥) + 𝑁(𝑥)𝐷(𝑥) − 𝑁(𝑥)𝐷(𝑥 + ℎ)
lim
𝐷2 (𝑥) ℎ→0
ℎ
=
𝐷(𝑥)(𝑁(𝑥 + ℎ) − 𝑁(𝑥)) − 𝑁(𝑥)(𝐷(𝑥 + ℎ) − 𝐷(𝑥))
1
lim
𝐷2 (𝑥) ℎ→0
ℎ
=
[𝐷(𝑥)lim
ℎ→0
𝐷2 (𝑥)
=
𝐷(𝑥)𝑁 ′ (𝑥) − 𝑁(𝑥)𝐷′ (𝑥)
𝐷2 (𝑥)
1
𝑁(𝑥 + ℎ) − 𝑁(𝑥)
𝐷(𝑥 + ℎ) − 𝐷(𝑥)
− 𝑁(𝑥)lim
]
ℎ→0
ℎ
ℎ
𝑥
If 𝑓(𝑥) = 𝑒 ⁄ 2 then 𝑁 = 𝑒 𝑥 and 𝐷 = 𝑥 2 and
𝑥
𝑓 ′ (𝑥) =
𝐷(𝑥)𝑁 ′ (𝑥) − 𝑁(𝑥)𝐷′ (𝑥) 𝑥 2 𝑒 𝑥 − 𝑒 𝑥 (2𝑥) 𝑥(𝑥 − 2)𝑒 𝑥 (𝑥 − 2)𝑒 𝑥
=
=
=
𝐷2 (𝑥)
𝑥4
𝑥4
𝑥3
Example 5. Find the derivative of 𝑓(𝑥) =
𝑥
4
𝑥−3
3
𝐷(𝑥)𝑁 ′ (𝑥) − 𝑁(𝑥)𝐷′ (𝑥) (𝑥 − 3)(1) − 𝑥(1)
3
𝑓 ′ (𝑥) =
=
=−
2
2
(𝑥
(𝑥
𝐷 (𝑥)
− 3)
− 3)2
2
f(x)
1
f'(x)
0
-10
-5
-1
-2
0
5
10
Example 6. Find the derivative of 𝑓(𝑥) =
𝑓 ′ (𝑥) =
𝐷(𝑥)𝑁
′ (𝑥)
𝑥 2 +1
20
2𝑥−3
′
15
(𝑥 2
10
2
− 𝑁(𝑥)𝐷 (𝑥) (2𝑥 − 3)(2𝑥) −
+ 1)(2) 2𝑥 − 6𝑥 − 2
=
=
(2𝑥 − 3)2
(2𝑥 − 3)2
𝐷2 (𝑥)
5
0
-4
-2 -5 0
2
4
f(x)
6
f'(x)
-10
-15
-20
-25
-30
Example 7. Find the derivative of 𝑓(𝑥) =
ln 𝑥
1.4
1+𝑥
1.2
1
1
1
𝐷(𝑥)𝑁 ′ (𝑥) − 𝑁(𝑥)𝐷′ (𝑥) (1 + 𝑥) 𝑥 − (ln 𝑥)(1) 𝑥 + 1 − ln 𝑥
′
𝑓 (𝑥) =
=
=
(1 + 𝑥)2
(1 + 𝑥)2
𝐷2 (𝑥)
0.8
0.6
f(x)
0.4
f'(x)
0.2
0
-0.2 0
5
10
-0.4
Example 8. The number of CDs sold is given by 𝑁(𝑡) =
90𝑡 2
𝑡 2 +50
where N is thousands of CDs and t is the number of
months since the CD was released.
Total CD Sales
2
𝑁(10) =
90∙10
102 +50
= 60 thousand CDs will be sold in the first 10 months.
(𝑡 2 + 50)180𝑡 − 90𝑡 2 (2𝑡)
9000𝑡
𝑁 ′ (𝑡) =
= 2
(𝑡 2 + 50)2
(𝑡 + 50)2
Thousands of CDs
A) Find N(10) and N’(10) and interpret the results.
100
80
60
40
20
0
N(t)
N'(t)
0
4
8
12
16
20
24
Months after Release
𝑁 ′ (10)
=
9000∙10
(102 +50)2
= 4, total sales are increasing at the rate of 4,000 CD’s per
month.
B) Use the results of part A to estimate the total sales after 11 months. Compare this estimate to the actual value.
𝑁(11) ≈ 𝑁(10) + 𝑁 ′ (10) = 60 + 4 = 64 thousand CDs will be sold in the first 11 months.
𝑁(11) =
90∙112
112 +50
= 63,684 thousand CDs will be sold in the first 11 months. The estimated value was 316 CDs higher
than the actual number of CDs (an error of about 0.5%).
Example 9. A drug is injected into the bloodstream of a patient through the right arm. The concentration of the
drug (in milligrams per cubic centimeter) in the blood stream of the left arm t hours after the injection is given by
𝐶(𝑡) =
0.14𝑡
𝑡 2+1
Drug Concentration
2
𝐶 ′ (𝑡) =
(𝑡 + 1)(0.14) − 0.14𝑡(2𝑡)
( 𝑡 2 + 1)
𝐶 ′ (0.5) =
0.14(1−0.52 )
2
(0.5 +1)
2
2
=
0.14(1 − 𝑡
( 𝑡 2 + 1)
2)
0.15
0.1
mg/cc
A) Find the instantaneous rate of change in the concentration of
the drug relative to time after a half hour and after 3 hours.
Interpret the results.
2
C(t)
0.05
C'(t)
0
0
-0.05
2
4
6
Hours
= 0.0672. The concentration is increasing at
the rate of 0.0672 mg/cc per hour.
𝐶 ′ (3) =
0.14(1−32 )
(32 +1)
2
− 0.0112. The concentration is decreasing at the rate of 0.0112 mg/cc per hour.
B) What is the highest concentration of the drug and when does it occur?
𝐶 ′ (𝑡) =
0.14(1−𝑡 2 )
(𝑡2 +1)
2
= 0 when t = 1 hour. The highest concentration is 𝐶(1) =
0.14∙1
12 +1
= 0.07 mg/cc.
8
10
12