Difficulty: Intermediate | Prerequisites: Logarithmic properties (Part 1 of these notes), basic derivative rules (power rule, limit definition)
This is the second half of Chapter 4, where logarithms stop being abstract and start doing work. Continuous compounding uses the formula Pe^(rt) to model investments that grow without discrete compounding periods, and solving for time in that formula requires the natural log skills from Part 1. The chapter then proves that e^x is its own derivative, a result that makes exponential functions uniquely clean to differentiate, and walks through product-rule and quotient-rule applications with e^x. If you are sitting the exam, this section is where most of the marks live.
Continuous compounding uses A = Pe^(rt) to model growth where interest is applied every instant, and solving for how long it takes an investment to double boils down to t = ln(2)/r. The derivative of e^x is e^x (the only function that equals its own derivative), which makes differentiating exponential expressions a matter of combining that fact with the product rule or quotient rule.
Continuous compounding
A model of growth where interest is compounded infinitely often, rather than monthly, quarterly, or annually. The formula is A = Pe^(rt), where P is the principal, r is the annual rate, t is time in years, and A is the accumulated amount.
Think of it as the theoretical limit of "what if the bank added interest every second, every millisecond, every instant?"
Doubling time
The time it takes for a continuously compounded investment to double in value. Derived by setting A = 2P and solving: t = ln(2)/r.
In simple terms, this tells you how many years until your money doubles, and the formula works for any interest rate.
e (Euler's number)
The mathematical constant approximately equal to 2.71828. Defined as the limit: e = lim (n approaches infinity) of (1 + 1/n)^n.
Think of it as the number that naturally emerges when you compound interest more and more frequently. It is not chosen arbitrarily; it is the base that makes calculus with exponentials clean.
Derivative of e^x
The derivative of e^x is e^x. This is the defining property of the exponential function: it is the only function whose rate of change at every point equals the function's value at that point.
In simple terms, e^x grows at a rate exactly equal to its current size.
Product rule (with exponentials)
Used when differentiating a product of two functions, one or both involving e^x. If f(x) = u(x) * v(x), then f'(x) = u'(x)*v(x) + u(x)*v'(x).
Quotient rule (with exponentials)
Used when differentiating a fraction where e^x appears in the numerator, denominator, or both. If f(x) = u(x)/v(x), then f'(x) = [u'(x)*v(x) - u(x)*v'(x)] / [v(x)]^2.
Problem: If you invest $1,000 at an 8% annual interest rate, compounded continuously, how long does it take to double?
Start with the formula: 1000 * e^(0.08t) = 2000
Divide both sides by 1000: e^(0.08t) = 2
Take ln of both sides: ln(e^(0.08t)) = ln(2)
Simplify using ln(e^x) = x: 0.08t = ln(2)
Solve for t: t = ln(2) / 0.08, which is approximately 8.66 years
Problem: If you invest $P at an annual rate r, compounded continuously, how long to double?
Pe^(rt) = 2P
e^(rt) = 2 (the P cancels)
rt = ln(2)
t = ln(2) / r
Note: as r approaches 0, ln(2)/r approaches infinity. A near-zero interest rate means an extremely long wait to double, which makes intuitive sense.
The number e is defined as: e = lim (h approaches 0) of (1 + h)^(1/h).
Equivalently, with the substitution n = 1/h: e = lim (n approaches infinity) of (1 + 1/n)^n.
This limit arises naturally from asking "what happens if we compound interest more and more frequently?" The answer converges to e, rather than growing without bound.
This proof uses the limit definition of the derivative.
Start: d/dx(e^x) = lim (h approaches 0) of [e^(x+h) - e^x] / h
Factor: = lim (h approaches 0) of [e^x * e^h - e^x] / h
Factor out e^x: = e^x * lim (h approaches 0) of [e^h - 1] / h
The key step is showing that lim (h approaches 0) of [e^h - 1] / h = 1. The notes derive this through a substitution:
Let y = e^h - 1, so e^h = y + 1, and h = ln(y + 1)
As h approaches 0, y approaches 0
The limit becomes: lim (y approaches 0) of y / ln(y + 1)
Rewrite: lim (y approaches 0) of 1 / [(1/y) * ln(y + 1)]
Which equals: lim (y approaches 0) of 1 / ln((1 + y)^(1/y))
As y approaches 0, (1 + y)^(1/y) approaches e (by definition)
So the limit becomes: 1 / ln(e) = 1 / 1 = 1
Therefore: d/dx(e^x) = e^x * 1 = e^x.
e^x is the only function whose derivative equals itself. This single fact is why e is the preferred base for exponential functions in calculus.
Example 1 (product rule): f(x) = x^2 * e^x
u = x^2, so u' = 2x
v = e^x, so v' = e^x
f'(x) = 2x * e^x + e^x * x^2 = e^x(2x + x^2)
Example 2 (quotient rule): f(x) = e^x * 2 / (e^x + 2), which can be written as f(x) = (x^3) / (e^x + 2) based on the lecture notes
From the notes, the specific example given is f(x) = x^3 / (e^x + 2):
u = x^3, so u' = 3x^2
v = e^x + 2, so v' = e^x
f'(x) = [3x^2 * (e^x + 2) - e^x * x^3] / (e^x + 2)^2
Formula | Expression | When to use it |
|---|---|---|
Continuous compounding | A = Pe^(rt) | Finding future value with continuous growth |
Doubling time | t = ln(2) / r | Finding how long to double at rate r |
Definition of e | e = lim (1 + 1/n)^n as n approaches infinity | Understanding where e comes from |
Derivative of e^x | d/dx(e^x) = e^x | Any time you differentiate e^x |
Product rule | d/dx[u * v] = u'v + uv' | Differentiating products like x^2 * e^x |
Quotient rule | d/dx[u/v] = (u'v - uv') / v^2 | Differentiating fractions with e^x |
The doubling-time formula t = ln(2)/r is one of the most practical results in this course. Financial advisors use it to give clients a quick estimate: at 7% annual return, your money doubles in roughly 10 years (ln(2)/0.07 is about 9.9).
The fact that e^x is its own derivative is not just a mathematical curiosity. It is the reason exponential models are used for population growth, radioactive decay, drug metabolism, and the spread of diseases. In each case, the rate of change is proportional to the current amount, and e^x is the function that captures that behaviour exactly.
Students often think the derivative of e^x is x * e^(x-1), applying the power rule by mistake. The power rule applies to x^n, not to e^x. The derivative of e^x is simply e^x.
Students sometimes forget to apply the chain rule when the exponent is more complex. The derivative of e^(3x) is 3e^(3x), not e^(3x). Whenever the exponent is anything other than plain x, the chain rule is needed.
Students sometimes confuse the doubling-time formula. The correct formula is t = ln(2)/r, not t = 2/r or t = ln(r)/2. The ln(2) in the numerator comes from taking the natural log of 2 when you set e^(rt) = 2.
Students occasionally think that because d/dx(e^x) = e^x, the derivative of 2^x must be 2^x. It is not. The derivative of a^x (for any base a) is a^x * ln(a). The "equals itself" property is unique to base e.
⚠️ The doubling-time derivation (setting Pe^(rt) = 2P and solving for t) is a classic exam question. Be ready to do it from scratch, not just recall the formula.
⚠️ You may be asked to explain why d/dx(e^x) = e^x, or at least outline the limit-based proof. Know the key substitution step where lim (h approaches 0) of (e^h - 1)/h = 1.
⚠️ Product-rule and quotient-rule problems with e^x are standard exam fare. Practise factoring e^x out of the result, as examiners often want a simplified final form.
⚠️ The definition of e as a limit (lim (1 + 1/n)^n) may appear as a conceptual question or as a step in a derivation. Know it cold.
True or false: The derivative of e^(5x) is e^(5x). Answer: False. The chain rule gives 5e^(5x).
Fill in the blank: The doubling time for a continuously compounded investment at rate r is t = ___. Answer: ln(2)/r.
True or false: e^x is the only function whose derivative equals itself. Answer: True (up to constant multiples; Ce^x also works, but e^x is the canonical case).
Fill in the blank: lim (n approaches infinity) of (1 + 1/n)^n = ___. Answer: e.
True or false: The derivative of x^3 * e^x requires the quotient rule. Answer: False. It requires the product rule. The quotient rule is for fractions.
Q: You invest $5,000 at 6% annual interest, compounded continuously. How long until the investment doubles?
A: t = ln(2) / 0.06, which is approximately 11.55 years. Set 5000e^(0.06t) = 10000, divide by 5000, take ln of both sides, solve for t.
Q: Find f'(x) for f(x) = x * e^x.
A: Using the product rule: f'(x) = 1 * e^x + x * e^x = e^x(1 + x).
Q: Find f'(x) for f(x) = e^x / (x^2 + 1).
A: Using the quotient rule: f'(x) = [e^x(x^2 + 1) - e^x(2x)] / (x^2 + 1)^2 = e^x(x^2 - 2x + 1) / (x^2 + 1)^2. The numerator factors further as e^x(x - 1)^2.
Q: Outline the key step in proving d/dx(e^x) = e^x.
A: After factoring e^x out of the limit definition, you need to show that lim (h approaches 0) of (e^h - 1)/h = 1. This is done by substituting y = e^h - 1, converting h to ln(y + 1), and using the definition of e to evaluate the resulting limit.
Q: What is the doubling time for an investment at 100% annual interest, compounded continuously?
A: t = ln(2) / 1 = ln(2), which is approximately 0.693 years, or about 8.3 months. Even at 100% interest, continuous compounding does not double the investment instantly.
The logarithm properties from Part 1 of these notes are the tool you reach for every time you solve a continuous compounding equation. The two halves of Chapter 4 are tightly linked: logs let you undo what e^x does.
The result d/dx(e^x) = e^x feeds directly into later chapters on optimisation. When you set a derivative equal to zero to find maxima and minima, expressions involving e^x simplify cleanly because the exponential part never vanishes (e^x is always positive), so the critical points come from the other factor.
In Calculus II, knowing that e^x is its own derivative also means it is its own antiderivative, which simplifies integration of exponential functions considerably.
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