Difficulty: Intermediate to Advanced | Prerequisites: Sections 2.1–2.7 (limits, basic derivatives, power rule)
Exponential and logarithmic functions model growth and decay in finance (compound interest), biology (population growth), and pharmacology (drug concentration). The chain rule is the tool that lets you differentiate compositions of functions, which appear constantly once you move past the simplest examples. Combined with the product and quotient rules, it handles every function you will meet in this course. Logarithmic differentiation is a technique for complex products and quotients that would be painful to differentiate by brute force.
Compound interest uses A = P(1 + r/n)^(nt) for periodic compounding and A = Pe^(rt) for continuous. The derivative of e^(u) is e^(u) · u' (chain rule), and the derivative of ln(u) is u'/u. The product rule handles f · g, the quotient rule handles f/g, and the chain rule handles f(g(x)). For ugly products and quotients, take the natural log of both sides first (logarithmic differentiation).
Compound interest (periodic)
A = P(1 + r/n)^(nt), where P = principal, r = annual rate (decimal), n = compounding periods per year, t = years. In simple terms, the interest you earned last period starts earning its own interest next period.
Continuous compounding
A = Pe^(rt). The limit of compound interest as the compounding frequency goes to infinity. Used whenever a problem says "compounded continuously."
Exponential growth/decay model
Q(t) = Q₀ e^(kt). If k > 0, growth; if k < 0, decay. Used for population, radioactive decay, and similar problems.
Chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In simple terms, differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Product rule
d/dx [f(x) · g(x)] = f'(x) · g(x) + f(x) · g'(x). Used when two functions are multiplied together.
Quotient rule
d/dx [f(x)/g(x)] = [f'(x) · g(x) – f(x) · g'(x)] / [g(x)]². Used when one function is divided by another.
Logarithmic differentiation
A technique where you take ln of both sides, use log properties to simplify, then differentiate implicitly. Particularly useful for functions with products, quotients, and powers all mixed together.
For $6,500 at 3.6% for 15 years:
Annually (n=1): A = 6500(1 + 0.036/1)^(1·15) = 6500(1.036)^15 ≈ $11,075.40
Monthly (n=12): A = 6500(1 + 0.036/12)^(12·15) = 6500(1.003)^180 ≈ $11,147.06
Hourly (n=8760): A = 6500(1 + 0.036/8760)^(8760·15) ≈ $11,156.19
Continuously: A = 6500e^(0.036·15) = 6500e^(0.54) ≈ $11,156.22
The more frequently you compound, the more you earn, but the gains diminish quickly. Continuous compounding is the upper bound.
Model: P(t) = P₀ e^(kt). Given two data points, find k, then predict.
Example: Population was 10,000 in 2005 (t=0), 35,000 in 2021 (t=16). Find population in 2050 (t=45).
35,000 = 10,000 e^(16k). Divide: 3.5 = e^(16k). Take ln: ln(3.5) = 16k. So k = ln(3.5)/16 ≈ 0.0783.
P(45) = 10,000 e^(0.0783 · 45) = 10,000 e^(3.5235) ≈ 339,380.
d/dx [e^x] = e^x
d/dx [e^(u)] = e^(u) · u' (chain rule)
d/dx [ln(x)] = 1/x
d/dx [ln(u)] = u'/u (chain rule)
Example: y = ln(7e^(4x) / x⁵). Use log properties first: y = ln(7) + 4x – 5ln(x). Then dy/dx = 4 – 5/x.
Identify the outer and inner functions. Differentiate the outer, leave the inner alone, multiply by the derivative of the inner.
f(x) = ln(2x⁵ + 3x – 6). Outer: ln(u). Inner: u = 2x⁵ + 3x – 6. f'(x) = (10x⁴ + 3)/(2x⁵ + 3x – 6).
f(x) = 10(3x² + 5x)^(1/5). f'(x) = 10 · (1/5)(3x² + 5x)^(–4/5) · (6x + 5) = 2(6x+5)(3x²+5x)^(–4/5).
y = (x–2)⁵(4x+1)³. Use product rule, each factor needs chain rule.
dy/dx = 5(x–2)⁴ · (4x+1)³ + (x–2)⁵ · 3(4x+1)² · 4
= 5(x–2)⁴(4x+1)³ + 12(x–2)⁵(4x+1)²
Factor: (x–2)⁴(4x+1)²[5(4x+1) + 12(x–2)] = (x–2)⁴(4x+1)²(32x – 19).
f(x) = e^(3x) / (4x+5)⁵. Quotient rule + chain rule on each piece.
f'(x) = [3e^(3x)(4x+5)⁵ – e^(3x) · 5(4x+5)⁴ · 4] / (4x+5)¹⁰
= e^(3x)(4x+5)⁴ [3(4x+5) – 20] / (4x+5)¹⁰
= e^(3x)(12x – 5) / (4x+5)⁶.
Horizontal tangent: set numerator = 0. e^(3x) is never 0, so 12x – 5 = 0, x = 5/12.
For y = e^(2x)√(4x–1) / [x³(x+4)⁴]:
Take ln: ln(y) = 2x + (1/2)ln(4x–1) – 3ln(x) – 4ln(x+4).
Differentiate both sides: (1/y) · y' = 2 + 2/(4x–1) – 3/x – 4/(x+4).
Multiply by y: y' = y · [2 + 2/(4x–1) – 3/x – 4/(x+4)].
Substitute the original expression for y.
f(x) = 3x⁴ ln(x), domain (0, ∞).
f'(x) = 12x³ ln(x) + 3x⁴ · (1/x) = 12x³ ln(x) + 3x³ = 3x³(4ln(x) + 1).
Set f'(x) = 0: x³ = 0 gives x = 0 (not in domain). 4ln(x) + 1 = 0 gives ln(x) = –1/4, so x = e^(–1/4).
C(t) = 0.18t / (t² + 1). Use quotient rule:
C'(t) = [0.18(t²+1) – 0.18t · 2t] / (t²+1)² = 0.18(1 – t²) / (t²+1)².
C'(0.5) = 0.18(1 – 0.25)/(1.25)² = 0.18(0.75)/1.5625 ≈ 0.086. Concentration is increasing at 0.086 mg/cm³ per hour.
C'(4) = 0.18(1 – 16)/(17)² = 0.18(–15)/289 ≈ –0.009. Concentration is decreasing at 0.009 mg/cm³ per hour.
Compound interest: A = P(1 + r/n)^(nt)
Continuous compounding: A = Pe^(rt)
d/dx [e^(u)] = e^(u) · u'
d/dx [ln(u)] = u'/u
Product rule: (fg)' = f'g + fg'
Quotient rule: (f/g)' = (f'g – fg') / g²
Chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
Students often forget that d/dx [e^(3x)] is 3e^(3x), not just e^(3x). The chain rule multiplies by the derivative of the exponent.
Students often differentiate ln(x² + 1) as 1/(x² + 1) instead of 2x/(x² + 1). The chain rule gives the derivative of the inside in the numerator.
When using logarithmic differentiation, students forget to multiply by y at the end. The left side is (1/y)·y', so you must multiply both sides by y to isolate y'.
Students sometimes confuse the compounding formula (periodic) with the continuous formula. If the problem says "continuously," use Pe^(rt). Otherwise use P(1 + r/n)^(nt).
⚠️ Before differentiating ln of a fraction, simplify using log properties: ln(a/b) = ln(a) – ln(b). This avoids the quotient rule entirely.
⚠️ When a problem says "logarithmic differentiation," any other method scores zero.
⚠️ Compound interest problems ask you to show the setup. Write the formula with values plugged in before computing.
⚠️ Interpretation questions for drug concentration or revenue: state whether the quantity is increasing or decreasing, by how much, and at what time.
Fill in the blank: d/dx [e^(5x)] = __________. (5e^(5x))
True or False: d/dx [ln(x³)] = 3/x. (True. ln(x³) = 3ln(x), derivative = 3/x.)
Fill in the blank: For continuous compounding, the formula is A = __________. (Pe^(rt))
True or False: The product rule gives (fg)' = f' · g'. (False. It is f'g + fg'.)
Fill in the blank: In logarithmic differentiation, the first step is to take __________ of both sides. (the natural log / ln)
Q: Differentiate y = ln(7e^(4x) / x⁵).
A: Simplify first: y = ln(7) + 4x – 5ln(x). Then dy/dx = 4 – 5/x.
Q: Find where the tangent line is horizontal for f(x) = 3x⁴ ln(x) on (0, ∞).
A: f'(x) = 3x³(4ln(x) + 1) = 0. Since x > 0, set 4ln(x) + 1 = 0. ln(x) = –1/4. x = e^(–1/4).
Q: Find the tangent line to f(x) = 2e^(5x) – 3 at x = 0.
A: f(0) = 2e⁰ – 3 = –1. f'(x) = 10e^(5x). f'(0) = 10. Tangent line: y = 10x – 1.
Q: Find the x-value where the tangent is horizontal for f(x) = e^(3x)/(4x+5)⁵.
A: f'(x) = e^(3x)(12x–5)/(4x+5)⁶. Set numerator = 0. e^(3x) ≠ 0, so 12x – 5 = 0. x = 5/12.
Q: A town had population 10,000 in 2005 and 35,000 in 2021 (continuous growth). Estimate the population in 2050.
A: k = ln(3.5)/16 ≈ 0.0783. P(45) = 10,000 e^(0.0783·45) ≈ 339,380.
The chain, product, and quotient rules are used in every remaining chapter. The graphing strategy (Ch. 4) requires first and second derivatives of functions involving exponentials and logs. Optimisation problems (Ch. 4.5–4.6) use these derivative rules to find maximum and minimum values. Integration (Ch. 5) reverses these rules.
exponential function, e^x, natural log, ln, compound interest, continuous compounding, exponential growth, exponential decay, chain rule, product rule, quotient rule, logarithmic differentiation, derivative of e^x, derivative of ln(x), drug concentration, population growth model, horizontal tangent, MAC 2233, business calculus, FIU