Difficulty: Intermediate | Prerequisites: Natural logarithm properties (Part 1 of these notes), chain rule, product and quotient rules
This is the second half of Calc AB Unit 5. It covers three related areas: how to differentiate complicated products and quotients by taking the log of both sides first (logarithmic differentiation), the derivative and integral formulas for exponential and general logarithmic functions, and the exponential growth/decay model. These tools come together in applied problems, particularly in modelling populations, radioactive decay, and compound interest. If you are comfortable with ln properties and the chain rule, you are ready.
Logarithmic differentiation turns nasty products and quotients into manageable sums by taking ln of both sides. Exponential functions (e^x and a^x) have clean derivative and integral formulas. The growth/decay model A = Pe^(rt) is the standard framework for half-life, doubling time, and continuous growth problems.
Logarithmic differentiation
A technique where you take the natural log of both sides of y = f(x), use log properties to simplify, differentiate implicitly, then solve for dy/dx. Think of it as converting multiplication and division into addition and subtraction, which are far easier to differentiate.
Exponential function (base e)
The function e^x, which has the unique property that its derivative is itself: d/dx (e^x) = e^x. In simple terms, the rate of change of e^x at any point equals the value of e^x at that point.
General exponential function (base a)
The function a^x for any positive constant a ≠ 1. Its derivative introduces a factor of ln a: d/dx (a^x) = a^x · ln a. Think of ln a as a "conversion factor" that accounts for using a base other than e.
General logarithmic function (base a)
The function log_a(x), which is the inverse of a^x. Its derivative is d/dx log_a(x) = 1/(x · ln a). In simple terms, dividing by ln a converts from base-a to base-e.
Exponential growth/decay model
The formula A = Pe^(rt), where P is the initial amount, r is the continuous rate, and t is time. A positive r gives growth; a negative r gives decay. This is the go-to model for any quantity that changes at a rate proportional to its current size.
Half-life
The time it takes for a quantity to reduce to half its initial value under exponential decay. You find it by setting A = P/2 in the decay formula and solving for t.
These formulas appear constantly:
d/dx (e^x) = e^x
∫ e^u du = e^u + C
d/dx (a^x) = a^x · ln a
∫ a^x dx = a^x / ln a + C
d/dx log_a(u) = (1/(ln a)) · (1/u) · (du/dx), applying the chain rule
Find d/dx [log_3(x² – 3x)].
Apply the formula: d/dx log_a(u) = (1/(ln a)) · (1/u) · du/dx
Here a = 3, u = x² – 3x, du/dx = 2x – 3
Result: (2x – 3) / [ln 3 · (x² – 3x)]
The chain rule is doing the heavy lifting. The 1/ln 3 factor is just the base conversion.
Use this when you have a function that is a messy product, quotient, or has a variable in both the base and the exponent. The steps:
Take ln of both sides: ln y = ln[f(x)]
Use log properties to separate, then differentiate both sides (implicit differentiation on the left gives (1/y)(dy/dx))
Solve for dy/dx, substituting back the original expression for y if needed
Differentiate y = 2x(x – 2)² / √(x² + 1).
Step 1: Take ln of both sides.
ln y = ln[2x] + 2 ln(x – 2) – (1/2) ln(x² + 1)
Using the product, power, and quotient rules of logarithms.
Step 2: Differentiate both sides.
(1/y)(dy/dx) = 1/x + 2/(x – 2) – x/(x² + 1)
On the left, implicit differentiation gives (1/y)(dy/dx). On the right, each term differentiates using d/dx [ln u] = (1/u)(du/dx).
Step 3: Solve for dy/dx.
dy/dx = y · [1/x + 2/(x – 2) – x/(x² + 1)]
Substitute back the original expression for y:
dy/dx = [2x(x – 2)² / √(x² + 1)] · [1/x + 2/(x – 2) – x/(x² + 1)]
Without logarithmic differentiation, this would require the product rule nested inside the quotient rule, which is far messier.
The model: A = Pe^(rt)
A = amount at time t
P = initial amount (at t = 0)
r = continuous rate (positive for growth, negative for decay)
t = time
Finding the rate from a half-life:
If the half-life is known, set A = P/2 and solve for r:
P/2 = Pe^(r · t_half)
1/2 = e^(r · t_half)
ln(1/2) = r · t_half
r = –0.693 / t_half
Once you have r, you can solve for any unknown in the formula.
Thorium has a half-life of 25 days. How long will it take for a 256,000 g sample to decay to 1,000 g?
Step 1: Find r.
1/2 = e^(25r)
ln(1/2) = 25r
r = –0.693 / 25 = –0.027 (approximately)
Step 2: Solve for t.
1,000 = 256,000 · e^(–0.027t)
1,000 / 256,000 = e^(–0.027t)
ln(1/256) = –0.027t
t = ln(1/256) / (–0.027) ≈ 200 days
Formula | Notes |
|---|---|
d/dx (e^x) = e^x | Derivative equals itself |
∫ e^u du = e^u + C | With chain rule for u-sub |
d/dx (a^x) = a^x · ln a | The ln a factor handles the base |
∫ a^x dx = (1/ln a) · a^x + C | Divide by ln a |
d/dx log_a(u) = (1/(ln a)) · (1/u) · du/dx | Chain rule included |
A = Pe^(rt) | Continuous growth/decay |
r = –0.693 / t_half | Quick formula for decay rate from half-life |
Exponential decay models describe radioactive decay (as in the thorium example), drug metabolism in pharmacology, and the cooling of objects (Newton's law of cooling). Exponential growth models cover population dynamics, compound interest, and the spread of infectious diseases. Logarithmic differentiation is used in statistics when working with likelihood functions, since products of many terms become sums under the log.
Forgetting the ln a factor. When differentiating a^x, students often write the derivative as just a^x, dropping the ln a. The factor only disappears when a = e (because ln e = 1).
Sign errors in decay problems. The rate r is negative for decay. Students sometimes find a positive r and then wonder why their answer gives growth instead of decay.
Confusing half-life with the rate. The half-life is not r. You must solve for r using the half-life, then use r in the decay formula.
Skipping the "multiply by y" step in logarithmic differentiation. After differentiating ln y, students sometimes forget to multiply both sides by y (or substitute the original function back) to get the final answer for dy/dx.
⚠️ Logarithmic differentiation is a common free-response technique. If the problem gives a complicated product/quotient and asks for dy/dx, this is the expected method.
⚠️ The derivative formulas for e^x, a^x, and log_a(x) are tested directly on multiple-choice sections. Know them without hesitation.
⚠️ Growth and decay problems appear on nearly every AP exam. The two-step process (find r from given info, then solve for the unknown) is the standard approach.
⚠️ When doing logarithmic differentiation, always state the original function for y when substituting back. Leaving dy/dx in terms of y without substituting is marked as incomplete.
Fill in the blank: d/dx (a^x) = ______
A: a^x · ln a
True or false: The half-life of a substance is the same as its decay rate.
A: False. The half-life is the time for the quantity to halve. The decay rate r is found by r = –0.693 / t_half.
Fill in the blank: In logarithmic differentiation, the left side always becomes (1/y)(______) after differentiating.
A: dy/dx
True or false: If a substance has a half-life of 10 days, then r = –10.
A: False. r = –0.693/10 ≈ –0.069. The half-life and the rate are different quantities.
Q: Differentiate y = 5^x.
A: dy/dx = 5^x · ln 5. Apply the general exponential rule directly.
Q: Evaluate ∫ 3^x dx.
A: (1/ln 3) · 3^x + C. The integral of a^x is a^x / ln a + C.
Q: Find d/dx [log_2(5x)].
A: Using the formula: (1/(ln 2)) · (1/(5x)) · 5 = 1/(x · ln 2). The 5 in the numerator and denominator cancel.
Q: A bacteria population doubles every 4 hours. If you start with 500 bacteria, how many are there after 12 hours?
A: First find r: 2 = e^(4r), so r = ln 2 / 4 ≈ 0.173. Then A = 500 · e^(0.173 · 12) = 500 · e^(2.079) ≈ 4,000. (Equivalently, 12 hours is three doubling periods: 500 → 1,000 → 2,000 → 4,000.)
Q: Use logarithmic differentiation to find dy/dx if y = x^x.
A: Take ln of both sides: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + 1. So dy/dx = y(ln x + 1) = x^x(ln x + 1). This is a classic example where the variable appears in both the base and the exponent, making logarithmic differentiation the only clean approach.
The growth/decay model connects to differential equations, which you will study in later units (or in Calc BC). The equation dA/dt = rA has the solution A = Pe^(rt), so this unit is your first exposure to solving a differential equation. Logarithmic differentiation connects to implicit differentiation (Unit 3) since you are differentiating ln y with respect to x. The general log and exponential derivatives build on the chain rule from Unit 3 and prepare you for integration by parts in Calc BC.
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