1. Chapter 4 Additional Derivative Topics

2. Additional Derivative Topics
In this chapter, we will explore these topics:
Review exponential and log functions
Find derivatives of exponential and log functions
Find derivatives of products and quotients
Use the Chain Rule to find derivatives
Use Implicit Differentiation to find derivatives
Solve related rates problems
Find Elasticity of Demand
Barnett/Ziegler/Byleen Business Calculus 12e

3. Chapter 4 Additional Derivative Topics
Section 1
The Constant e and Continuous Compound Interest

4. Objectives for Section 4.1 e and Continuous Compound Interest
Students will review exponential and log applications involving e and ln
simple interest
compound interest
continuous compound interest
Barnett/Ziegler/Byleen Business Calculus 12e

5. The Constant e Reminder: By definition, e  2.718 281 828 459 …
e is defined as either one of the following limits:
Let’s verify these limits using our calculators…
𝑒= lim 𝑥→∞ 𝑥 𝑥
𝑒= lim 𝑥→ 𝑥 1 𝑥
Barnett/Ziegler/Byleen Business Calculus 12e

6. e  2.718 281 828 459 … 𝑦= 1+ 1 𝑥 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→∞, 𝑦→𝑒
𝑦= 𝑥 𝑥
𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒:
𝐴𝑠 𝑥→∞, 𝑦→𝑒
Barnett/Ziegler/Byleen Business Calculus 12e

7. e  2.718 281 828 459 … 𝑦= 1+𝑥 1 𝑥 𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒: 𝐴𝑠 𝑥→0, 𝑦→𝑒
𝑦= 1+𝑥 1 𝑥
𝐼𝑛 𝑇𝑅𝐴𝐶𝐸 𝑚𝑜𝑑𝑒:
𝐴𝑠 𝑥→0, 𝑦→𝑒
Barnett/Ziegler/Byleen Business Calculus 12e

8. Evaluating Expressions
Examples showing: e
6  e  e  9
𝑒 .05
𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑒 𝑥 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Evaluating Expressions

9. Interest Formulas P = principal (initial deposit)
r = annual interest rate t = time in years
n = number of time interest is compounded per year
A = amount in account at the end of the time period
Simple Interest:
Compound interest:
Continuously compounded interest:
𝐴=𝑃(1+𝑟𝑡)
𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡
𝐴=𝑃 𝑒 𝑟𝑡
Barnett/Ziegler/Byleen Business Calculus 12e

10. Simple vs Compound Interest
𝑆𝑖𝑚𝑝𝑙𝑒 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 − 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑
Year
Account Balance
Interest Earned 1
$1000
1000(.05) = $50 2
$1050 3
$1100 4
$1150
𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 − 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝒅𝒐𝒆𝒔 𝑖𝑛𝑐𝑙𝑢𝑑𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑
Year
Account Balance
Interest Earned 1
$1000
1000(.05) = $50 2
$1050
1050(.05) = $52.50 3 $
(.05) = $55.13 4 $
(.05) = $57.88
Barnett/Ziegler/Byleen Business Calculus 12e

11. Example 1
When you were born, your Grandma put $1,000 in a bank for you, at 5% interest. Calculate the amount in the account after 20 years using:
Simple interest
Interest compounded quarterly
Interest compounded continuously
(My Grandma)
Yuri Sasaki
1900−1997
Barnett/Ziegler/Byleen Business Calculus 12e

12. Example 1 (continued) Principal = $1,000 Rate = 5% Time = 20 years
Simple interest:
Compounded quarterly:
Compounded continuously:
𝐴= ∙20 =
$2000
𝐴= (20) = $
𝐴=1000 𝑒 .05(20) = $
Barnett/Ziegler/Byleen Business Calculus 12e

13. Example 2 IRA
After graduating from USC, Tammy Trojan lands a great job with Google. In her first year, she buys a $3,000 Roth IRA and invests it in a mutual fund that grows at 12% per year, compounded continuously. She plans to become semi-retired in 35 years.
What will be its value when she “retires”?
In her second year she buys an identical Roth IRA for $3,000. What will be its value in 34 years?
What advice can be learned from this, regarding when you should start saving for retirement?
Barnett/Ziegler/Byleen Business Calculus 12e

14. Example 2 (continued) Principal = $3000 Rate = 12% Time = 35 years
At the end of 35 years, the first account will have:
At the end of 34 years, the second account will have:
The lesson learned is once you have a full-time job, you should start saving for retirement as early as possible. Save a little every month and be amazed at how quickly it adds up!
𝐴=3000 𝑒 .12(35) =
$200,058.99
𝐴=3000 𝑒 .12(34) =
$177,436.41
Barnett/Ziegler/Byleen Business Calculus 12e

15. Example 3 Computing Growth Time
How long will it take an investment of $5,000 to grow to $8,000 if it is invested at 5% compounded continuously?
A=$8000 P=$5000 rate = 5% time=?
𝐴=𝑃 𝑒 𝑟𝑡
8000=5000 𝑒 .05𝑡
8 5 = 𝑒 .05𝑡
ln =𝑡
ln = ln 𝑒 .05𝑡
𝑡≈9.4
ln =.05𝑡∙ ln 𝑒
It will take about 9.4 years for the money to grow to $8000.
ln =.05𝑡
Barnett/Ziegler/Byleen Business Calculus 12e

16. Example 4
How long will it take an investment to triple if it is invested at 6.5% compounded every 4 months?
A=3 P=1 rate = 6.5% n = 3 time=?
𝑙𝑜𝑔 3 = 3𝑡∙𝑙𝑜𝑔
𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡
3= 𝑡
log 3 3 log =𝑡
𝑙𝑜𝑔 3 = 𝑙𝑜𝑔 𝑡
𝑡≈17.1
It will take about 17.1 years for the money to triple.
Barnett/Ziegler/Byleen Business Calculus 12e

17. Example 5
A CD will pay $50,000 at maturity 5 years from now. How much will you have to pay for the CD now if the money is 6.4% compounded continuously?
A=50,000 P=? rate = 6.4% time=5
𝐴=𝑃 𝑒 𝑟𝑡
50000=𝑃 𝑒 .064(5)
𝑒 .064(5) =𝑃
You will pay $36, for the CD in order for it to be worth $50,000 in 5 years.
𝑃≈$36,307.45
Barnett/Ziegler/Byleen Business Calculus 12e

18. Example 6
A family paid $99,000 for a house. Fifteen years later, the house sold for $195,000. If interest was compounded continuously, what annual rate of interest did the original investment earn?
A=195,000 P=99,000 rate = ? time=15
𝐴=𝑃 𝑒 𝑟𝑡
195000=99000 𝑒 𝑟(15)
𝑙𝑛 =𝑟
= 𝑒 15𝑟
𝑟≈.0452
𝑙𝑛 = ln 𝑒 15𝑟
The interest rate was 4.52%.
𝑙𝑛 = 15r∙ln 𝑒
Barnett/Ziegler/Byleen Business Calculus 12e

19. Homework
#4-1: Pg 215
(9, 13, 14, 17, 19,
21, 27, 29, 33, 35, 37)
Barnett/Ziegler/Byleen Business Calculus 12e