MAC 2233 Calculus for Business | Sections 3.4, 4.1, 4.2, 4.4 | Source: Exam 3 Review
This review covers every differentiation technique tested on Exam 3: the chain rule for composite functions, the product and quotient rules combined with the chain rule, derivatives of logarithmic and exponential functions, tangent line equations, and logarithmic differentiation. If you can identify the outer and inner function in a composition and apply the right rule (or combination of rules), you can handle every problem on this exam.
Difficulty: Intermediate
Prerequisites: Basic derivative rules (power rule, constant multiple rule, sum/difference rule), understanding of exponential and logarithmic functions, and familiarity with function composition from Sections 3.1 to 3.3.
Up to now, you have differentiated simple functions using the power rule, sum rule, and basic exponential/log rules. This unit asks: what happens when those simple pieces are nested inside each other, multiplied together, or divided? Sections 3.4 and 4.1 to 4.4 give you the chain rule, and then layer it on top of product, quotient, logarithmic, and exponential differentiation. Every applied calculus problem from here forward, whether it involves cost functions, revenue models, or growth rates, depends on combining these rules fluently.
Chain rule
The derivative of a composite function f(g(x)) is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function: d/dx [f(g(x))] = f'(g(x)) · g'(x).
In simple terms, differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Inner function / outer function
In a composite function f(g(x)), the inner function is g(x) (what you substitute in) and the outer function is f(u) (the larger structure wrapped around it).
Think of it as peeling layers: the outer function is the shell, the inner function is what is inside the shell.
Product rule
If h(x) = f(x) · g(x), then h'(x) = f'(x)·g(x) + f(x)·g'(x). Derivative of the first times the second, plus the first times the derivative of the second.
In simple terms, each factor gets a turn being differentiated while the other stays put, and you add the two results.
Quotient rule
If h(x) = f(x)/g(x), then h'(x) = [f'(x)·g(x) - f(x)·g'(x)] / [g(x)]². Low-d-high minus high-d-low, all over the square of what is below.
In simple terms, "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."
Logarithmic differentiation
A technique where you take the natural log of both sides of y = f(x), use log properties to simplify, then differentiate implicitly. Especially useful when the function is a product/quotient of many pieces raised to powers.
Think of it as turning a messy product into a tidy sum before differentiating.
Tangent line
The line that just touches a curve at a single point and has the same slope as the curve at that point. Its equation is y - f(a) = f'(a)(x - a).
In simple terms, the straight line that best approximates the curve right at that point.
Horizontal tangent line
A tangent line with slope zero, meaning f'(x) = 0 at that point. These occur at local maxima, local minima, or other flat spots on the graph.
In simple terms, where the curve momentarily levels out.
When to use it
Any time one function is plugged into another: (3x - 7)^13, e^(3x²+1), ln(4x³ - 2x + 1), (x² - 4x + 1)^(1/3).
If you can identify an "inside" piece and an "outside" wrapper, the chain rule applies.
The pattern
Differentiate the outer function, leaving the inner function untouched.
Multiply by the derivative of the inner function.
In Leibniz notation: if y = f(u) and u = g(x), then dy/dx = (dy/du) · (du/dx).
Worked example: f(x) = (3x - 7)^13
Outer function: u^13. Inner function: u = 3x - 7.
Derivative of outer: 13u^12. Derivative of inner: 3.
Result: f'(x) = 13(3x - 7)^12 · 3 = 39(3x - 7)^12.
Worked example: f(x) = 5(x² - 4x + 1)^(1/3)
Rewrite the cube root as a 1/3 power. Outer: 5u^(1/3). Inner: u = x² - 4x + 1.
f'(x) = 5 · (1/3)(x² - 4x + 1)^(-2/3) · (2x - 4) = 10(x - 2) / [3(x² - 4x + 1)^(2/3)].
Key habit: always ask, "What is the innermost piece that is not just x?" That is your inner function.
Product rule + chain rule
When a function is a product of two pieces and at least one of them requires the chain rule, apply the product rule first, then the chain rule within each derivative.
f(x) = x^5 · e^(3x²+1)
First factor: x^5, its derivative: 5x^4.
Second factor: e^(3x²+1), its derivative via chain rule: e^(3x²+1) · 6x.
Product rule: f'(x) = 5x^4 · e^(3x²+1) + x^5 · 6x · e^(3x²+1) = x^4 · e^(3x²+1)(5 + 6x²).
f(x) = x³ · ln(x³ + 2x - 3)
First: x³, derivative: 3x².
Second: ln(x³ + 2x - 3), derivative via chain rule: (3x² + 2)/(x³ + 2x - 3).
Product rule: f'(x) = 3x² ln(x³ + 2x - 3) + x³(3x² + 2)/(x³ + 2x - 3).
Quotient rule + chain rule
When a function is a fraction where the numerator or denominator (or both) needs the chain rule.
f(x) = 7x² / (2x + 3)³
Top: 7x², derivative: 14x.
Bottom: (2x + 3)³, derivative via chain rule: 3(2x + 3)² · 2 = 6(2x + 3)².
Quotient rule: f'(x) = [14x(2x + 3)³ - 7x² · 6(2x + 3)²] / (2x + 3)^6.
Factor and simplify: f'(x) = 14x(3 - x) / (2x + 3)^4.
f(x) = (x - 2)² / (3x + 1)
Apply the quotient rule, using the chain rule on the numerator.
f'(x) = [2(x - 2)(3x + 1) - (x - 2)² · 3] / (3x + 1)².
Factor out (x - 2): f'(x) = (x - 2)(3x + 8) / (3x + 1)².
When two separate product rules appear in one problem
Some functions are sums of two products. Differentiate each product separately using the product rule and add the results.
f(x) = x ln(1 - e^x) + x²(x + 1)³
First product: x · ln(1 - e^x). Apply the product rule; the chain rule kicks in when differentiating ln(1 - e^x), giving -e^x/(1 - e^x).
Second product: x² · (x + 1)³. Apply the product rule; the chain rule on (x + 1)³ gives 3(x + 1)².
Combine: f'(x) = ln(1 - e^x) - xe^x/(1 - e^x) + 2x(x + 1)³ + 3x²(x + 1)².n Tip: on the exam, simplify and factor wherever you can. The answer key often presents a fully factored form.
Natural logarithm derivative
d/dx [ln(x)] = 1/x.
With the chain rule: d/dx [ln(g(x))] = g'(x) / g(x).
The denominator is always the original inside function. The numerator is the derivative of that inside function.
Example: f(x) = ln(4x³ - 2x + 1). Inner function: 4x³ - 2x + 1. Its derivative: 12x² - 2.
f'(x) = (12x² - 2) / (4x³ - 2x + 1).
Exponential function derivative
d/dx [e^x] = e^x.
With the chain rule: d/dx [e^(g(x))] = e^(g(x)) · g'(x).
The exponential part reproduces itself; you just tack on the derivative of the exponent.
Example: f(x) = 5e^(x² - 4x + 1). The exponent is x² - 4x + 1, so its derivative is 2x - 4.
f'(x) = 5e^(x² - 4x + 1) · (2x - 4) = 10(x - 2)e^(x² - 4x + 1).
General exponential a^x
d/dx [a^x] = a^x · ln(a).
With the chain rule: d/dx [a^(g(x))] = a^(g(x)) · ln(a) · g'(x).
This appears in Problem 16 where the base is 10: d/dx [10^x] = 10^x · ln(10).
Chain rule inside a chain rule (nested compositions)
f(x) = (2x + ln(x))³ has an outer cube and an inner function 2x + ln(x). The derivative of the inner function is 2 + 1/x. So:
f'(x) = 3(2x + ln(x))² · (2 + 1/x).
Two distinct situations tested on this exam
The review distinguishes problems 13/14 from problems 15/16. Understanding this distinction is critical.
Situation A: the function already has ln on the outside (Problems 13, 14)
When f(x) = ln(something complicated), you expand using log properties first, then differentiate term by term. No implicit differentiation is needed because the left side is already a log.
ln(A/B) = ln(A) - ln(B)
ln(AB) = ln(A) + ln(B)
ln(A^n) = n · ln(A)
Example: f(x) = ln[x³ + 2x - 1] - ln[3x] - ln[(x + 1)^10]
Simplify: f(x) = ln(x³ + 2x - 1) - ln(3) - ln(x) - 10 ln(x + 1).
Differentiate each term: f'(x) = (3x² + 2)/(x³ + 2x - 1) - 1/x - 10/(x + 1).
Example: f(x) = ln[e^(2x) · (x³ - 2)^(1/2) / (x + 1)^4]
Expand: f(x) = 2x + (1/2)ln(x³ - 2) - 4 ln(x + 1).
Differentiate: f'(x) = 2 + 3x²/[2(x³ - 2)] - 4/(x + 1).
Note how ln(e^(2x)) simplifies directly to 2x, because ln and e are inverse functions.
Situation B: no ln on the outside, so you must take ln of both sides (Problems 15, 16)
When f(x) is a messy product/quotient of pieces raised to powers but there is no ln present, you:
Write y = f(x).
Take ln of both sides: ln(y) = ln(f(x)).
Expand the right side using log properties.
Differentiate both sides. The left side becomes (1/y) · y' by the chain rule.
Solve for y' by multiplying both sides by y (which is the original f(x)).
Example: f(x) = e^x · (x - 3)^10 / (x + 1)^4.
ln(y) = x + 10 ln(x - 3) - 4 ln(x + 1).
Differentiate: (1/y)·y' = 1 + 10/(x - 3) - 4/(x + 1).
Solve: f'(x) = [1 + 10/(x - 3) - 4/(x + 1)] · [e^x(x - 3)^10 / (x + 1)^4].
The key difference is that in Situation B, differentiating the left-hand side (ln y) produces the (1/y)·y' term. That is the whole point of logarithmic differentiation.
The recipe
Find f(a) to get the y-coordinate of the point.
Find f'(x), then evaluate f'(a) to get the slope.
Plug into point-slope form: y - f(a) = f'(a)(x - a).
Example: y = ln(1 - x² + 2x⁴) at x = 1.
f(1) = ln(1 - 1 + 2) = ln(2).
f'(x) = (-2x + 8x³)/(1 - x² + 2x⁴). At x = 1: f'(1) = (-2 + 8)/(1 - 1 + 2) = 6/2 = 3.
Tangent line: y - ln(2) = 3(x - 1), which gives y = 3x - 3 + ln(2).
Example: f(x) = 5e^(x² - 4x + 1) at x = 0.
f(0) = 5e^(0 - 0 + 1) = 5e.
f'(x) = 5e^(x² - 4x + 1) · (2x - 4). At x = 0: f'(0) = 5e · (-4) = -20e.
Tangent line: y = -20ex + 5e.
Finding horizontal tangent lines
Set f'(x) = 0 and solve for x.
For f(x) = (2x - 3)^8 · (5x - 2)^7, use the product rule with chain rule:
f'(x) = (2x - 3)^7 · (5x - 2)^6 · (150x - 137).
Set each factor equal to zero: x = 3/2, x = 2/5, x = 137/150.
All three values give horizontal tangent lines.
Rule | Formula |
|---|---|
Chain rule | d/dx [f(g(x))] = f'(g(x)) · g'(x) |
Product rule | d/dx [f · g] = f' · g + f · g' |
Quotient rule | d/dx [f/g] = (f'g - fg') / g² |
ln derivative | d/dx [ln(g(x))] = g'(x) / g(x) |
e^x derivative | d/dx [e^(g(x))] = e^(g(x)) · g'(x) |
General exponential | d/dx [a^(g(x))] = a^(g(x)) · ln(a) · g'(x) |
Tangent line | y - f(a) = f'(a)(x - a) |
Log properties | ln(AB) = ln A + ln B; ln(A/B) = ln A - ln B; ln(A^n) = n ln A |
The chain rule is what lets you differentiate any model where one quantity depends on another that itself changes. In business, a revenue function R(x) = x · p(x) where price depends on demand means you need the product rule; if price is a nonlinear function of demand (like p = 2600 - 0.26x²), you need the chain rule when differentiating revenue or profit. Logarithmic and exponential derivatives show up in compound interest models, population growth, and drug concentration problems.
Students often forget to multiply by the derivative of the inner function when applying the chain rule. The derivative of (3x - 7)^13 is not 13(3x - 7)^12. It is 13(3x - 7)^12 · 3 = 39(3x - 7)^12. Missing that final factor of 3 is the single most common error.
Students confuse the product rule with the chain rule. If two functions are multiplied together, that is a product rule problem. If one function is plugged inside another, that is the chain rule. The fact that both may appear in the same problem does not make them interchangeable.
When differentiating ln(g(x)), students sometimes write g'(x) · ln(g(x)) as though the ln follows the same pattern as e^x. It does not. The derivative of ln(g(x)) is g'(x)/g(x), not g'(x) · ln(g(x)).
In logarithmic differentiation (Situation B), students forget to multiply by y = f(x) at the end. The whole point of the method is that (1/y)·y' appears on the left side, so you must solve for y' by multiplying through by y.
⚠️ Problems 1 to 9 test pure differentiation. Expect at least half of the exam to be "differentiate this function" problems requiring the chain rule combined with product, quotient, or logarithmic rules.
⚠️ Tangent line problems (10, 11) require you to evaluate both f(a) and f'(a) correctly. An arithmetic slip in either one costs you the entire problem.
⚠️ Problem 12 (horizontal tangent) tests your ability to factor f'(x) completely and set each factor to zero. This type of problem is a favourite exam question.
⚠️ The distinction between problems 13/14 and 15/16 is explicitly flagged in the review. Expect the exam to include at least one of each type. Know which situation calls for taking ln of both sides and which does not.
True or False: The derivative of ln(5x² + 1) is (10x) / (5x² + 1). True.
True or False: The derivative of e^(4x) is e^(4x). False. It is 4e^(4x) – you must multiply by the derivative of the exponent.
Fill in the blank: When using logarithmic differentiation, differentiating ln(y) with respect to x gives _____. Answer: (1/y) · y'.
True or False: To find a horizontal tangent line, you set f(x) = 0. False. You set f'(x) = 0.
Fill in the blank: ln(A^n) = _____. Answer: n · ln(A).
Q: Differentiate f(x) = (3x - 7)^13.
A: f'(x) = 39(3x - 7)^12. Apply the chain rule: bring down the 13, reduce the power by one, multiply by the derivative of (3x - 7), which is 3.
Q: Differentiate f(x) = x^5 · e^(3x² + 1).
A: f'(x) = x^4 · e^(3x² + 1) · (5 + 6x²). Product rule first, then chain rule on the exponential. Factor out common terms.
Q: Find the equation of the tangent line to y = ln(1 - x² + 2x⁴) at x = 1.
A: y = 3x - 3 + ln(2). Evaluate f(1) = ln(2), compute f'(1) = 3, then use point-slope form.
Q: What is the difference between problems where f(x) = ln(something) and problems where you use logarithmic differentiation?
A: When f(x) already equals ln(something), expand using log properties and differentiate directly. When f(x) is a product/quotient with no ln, take ln of both sides first, expand, differentiate (getting (1/y)y' on the left), then solve for y' by multiplying by y = f(x).
Q: Find the x-values where the tangent line to f(x) = (2x - 3)^8(5x - 2)^7 is horizontal.
A: x = 3/2, x = 2/5, x = 137/150. Set f'(x) = 0. Factor f'(x) completely: (2x - 3)^7(5x - 2)^6(150x - 137) = 0, then solve each factor.
Q: Differentiate f(x) = 7x² / (2x + 3)³.
A: f'(x) = 14x(3 - x) / (2x + 3)^4. Quotient rule with chain rule on the denominator, then factor and simplify.
The chain rule is the foundation for everything that follows in this course. Part 2 of these notes (increasing/decreasing intervals and local extrema) requires you to find f'(x) first, which means applying the techniques from this section. Part 3 (concavity and graphing strategy) goes one step further by differentiating f'(x) again. If your differentiation is shaky, every subsequent topic becomes harder. The applied business problems (cost, revenue, profit) are just differentiation problems wrapped in a word-problem context.
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