Difficulty: Intermediate to Advanced | Prerequisites: Power rule, derivatives of ln(x) and e^x, tangent line method
The product rule and quotient rule let you differentiate functions that are built from two pieces multiplied or divided together. This is where the exam gets heavier: you combine these rules with the derivative rules for ln(x) and e^x you have already learnt, and you apply them to tangent lines, horizontal tangents and word problems involving sales, subscribers and profit. If you are comfortable with the individual derivative rules, the remaining work is mechanical, pattern-recognition and simplification.
Product rule: (fg)' = f'g + fg'. Quotient rule: (f/g)' = (f'g - fg') / g². Use them whenever two functions are multiplied or divided. Then simplify, factor where possible, and apply to tangent lines or marginal-analysis problems as needed.
Product rule
If y = f(x) * g(x), then y' = f'(x) * g(x) + f(x) * g'(x). In words: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If y = f(x) / g(x), then y' = [f'(x) * g(x) - f(x) * g'(x)] / [g(x)]². In words: "low d-high minus high d-low, over the square of what's below."
Think of it as a formula where the denominator always gets squared, and the numerator follows a subtraction pattern.
Horizontal tangent
A tangent line with slope zero. Found by setting f'(x) = 0 and solving for x. The function has a local maximum, minimum or inflection point at these x-values.
Marginal analysis with rational functions
When a total function (sales, subscribers) is a fraction of polynomials, its derivative requires the quotient rule. The derivative then serves the same purpose as before: estimating the change from one period to the next.
Problem 26: f(x) = 5x²(x³ + 2). Let u = 5x², v = x³ + 2. u' = 10x, v' = 3x². f'(x) = 10x(x³ + 2) + 5x²(3x²) = 10x⁴ + 20x + 15x⁴ = 25x⁴ + 20x.
Problem 27: y = 5x⁸ * e^x. u = 5x⁸, v = e^x. u' = 40x⁷, v' = e^x. y' = 40x⁷ * e^x + 5x⁸ * e^x = 5x⁷ * e^x(8 + x) = 5x⁷ * e^x(x + 8).
Problem 28: f(x) = x⁷ * ln(x). u = x⁷, v = ln x. u' = 7x⁶, v' = 1/x. f'(x) = 7x⁶ * ln x + x⁷ * (1/x) = 7x⁶ ln x + x⁶ = x⁶(7 ln x + 1) = x⁶(1 + 7 ln x).
Problem 33: y = (1 + x - x²) * e^x. u = 1 + x - x², v = e^x. u' = 1 - 2x, v' = e^x. y' = (1 - 2x)e^x + (1 + x - x²)e^x = e^x[(1 - 2x) + (1 + x - x²)] = e^x(2 - x - x²).
Problem 29: y = x² / (2x - 1). f = x², g = 2x - 1. f' = 2x, g' = 2. y' = [2x(2x - 1) - x²(2)] / (2x - 1)² = [4x² - 2x - 2x²] / (2x - 1)² = (2x² - 2x) / (2x - 1)².
Problem 30: y = (t³ - 3t) / (t² - 4). f = t³ - 3t, g = t² - 4. f' = 3t² - 3, g' = 2t. y' = [(3t² - 3)(t² - 4) - (t³ - 3t)(2t)] / (t² - 4)². Expand the numerator: (3t⁴ - 12t² - 3t² + 12) - (2t⁴ - 6t²) = 3t⁴ - 15t² + 12 - 2t⁴ + 6t² = t⁴ - 9t² + 12. Final: y' = (t⁴ - 9t² + 12) / (t² - 4)².
Problem 31: f(x) = ln(x) / (1 + x). f = ln x, g = 1 + x. f' = 1/x, g' = 1. f'(x) = [(1/x)(1 + x) - ln(x)(1)] / (1 + x)². Simplify the numerator: (1 + x)/x - ln x = (1 + x - x ln x) / x. So f'(x) = (1 + x - x ln x) / [x(1 + x)²].
Problem 34: f(x) = (x - 8)/(3x - 4) at x = 2. f(2) = (2 - 8)/(6 - 4) = -6/2 = -3. f'(x) = [(1)(3x - 4) - (x - 8)(3)] / (3x - 4)² = [3x - 4 - 3x + 24] / (3x - 4)² = 20 / (3x - 4)². f'(2) = 20 / 4 = 5. Tangent: y - (-3) = 5(x - 2), so y = 5x - 13.
Problem 36: f(x) = (2x - 3)(x² - 6). Use the product rule: f'(x) = 2(x² - 6) + (2x - 3)(2x) = 2x² - 12 + 4x² - 6x = 6x² - 6x - 12. Set f'(x) = 0: 6x² - 6x - 12 = 0, divide by 6: x² - x - 2 = 0, factor: (x - 2)(x + 1) = 0. So x = 2 and x = -1.
Problem 32: S(t) = 150t / (t + 3). S'(t) = [150(t + 3) - 150t(1)] / (t + 3)² = 450 / (t + 3)². S(12) = 1800/15 = 120, meaning $120,000. S'(12) = 450/225 = 2, meaning $2,000 in new sales during month 13. Estimated total after 13 months: S(12) + S'(12) = 120 + 2 = 122, or $122,000.
Problem 35: N(t) = 90t² / (t² + 50). N'(t) = [180t(t² + 50) - 90t²(2t)] / (t² + 50)² = [180t³ + 9000t - 180t³] / (t² + 50)² = 9000t / (t² + 50)². N(10) = 9000/150 = 60 (60,000 subscribers). N'(10) = 90000/22500 = 4 (4,000 new subscribers in month 11). Estimated total: 60 + 4 = 64, or 64,000.
Problem 37: Demand S(x) = 100/x², cost C(x) = 0.5x² - 2x + 100. Revenue = price * quantity = x * (100/x²) = 100/x. Profit P(x) = 100/x - (0.5x² - 2x + 100) = 100/x - 0.5x² + 2x - 100. P'(x) = -100/x² - x + 2. P'(5) = -100/25 - 5 + 2 = -4 - 5 + 2 = -7. The profit is decreasing at $7 per unit increase in price when the price is $5.
(fg)' = f'g + fg' \quad \text{(product rule)}\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \quad \text{(quotient rule)}Horizontal tangent: set f'(x) = 0 and solve.
Swapping the sign in the quotient rule. The numerator is f'g minus fg', not the other way round. One reliable mnemonic: "low d-high minus high d-low" (denominator times derivative of numerator, minus numerator times derivative of denominator).
Forgetting to square the denominator. The quotient rule denominator is always [g(x)]², not g(x).
Not simplifying after applying the rule. The exam expects a simplified answer. Factor out common terms where possible (e.g. x⁶, e^x).
Using the quotient rule when you could simplify first. If f(x) = 5x²(x³ + 2), you could expand to 5x⁵ + 10x and use the power rule. The product rule also works, but either way, recognise your options.
Setting f(x) = 0 instead of f'(x) = 0 for horizontal tangents. The tangent is horizontal where the derivative is zero, not where the function is zero.
Arithmetic errors in the numerator expansion. Quotient rule problems often require expanding a product of two binomials in the numerator. Take it slowly and collect like terms carefully.
⚠️ Problems 26-31 are pure differentiation drills. Speed and accuracy here earn easy marks. Practise until the product and quotient rules are automatic.
⚠️ Problem 33 combines the product rule with e^x. Factor e^x out at the end for a clean answer.
⚠️ Tangent line at a specific point (Problem 34) requires the quotient rule to find f'(x), then evaluating at that point. Show all steps.
⚠️ Horizontal tangent (Problem 36) requires the product rule, setting f'(x) = 0, and solving the resulting quadratic. Know how to factor.
⚠️ Applied problems (32, 35) follow the same pattern: find S'(t) or N'(t) using the quotient rule, evaluate at a specific t, and interpret. Then add: total at (t+1) is approximately f(t) + f'(t).
Fill in the blank: If y = f(x) * g(x), then y' = ___. (f'g + fg')
True or False: The quotient rule numerator is f'g + fg'. (False. It is f'g - fg'. The product rule uses plus; the quotient rule uses minus.)
Fill in the blank: A horizontal tangent occurs where f'(x) = ___. (0)
True or False: If S(12) = 120 and S'(12) = 2, then S(13) is approximately 122. (True.)
Fill in the blank: In the quotient rule, the denominator of the result is ___. ([g(x)]²)
Q: Differentiate f(x) = 5x²(x³ + 2).
A: Product rule: f'(x) = 10x(x³ + 2) + 5x²(3x²) = 10x⁴ + 20x + 15x⁴ = 25x⁴ + 20x.
Q: Differentiate y = 5x⁸ * e^x.
A: y' = 40x⁷ * e^x + 5x⁸ * e^x = 5x⁷e^x(x + 8).
Q: Differentiate f(x) = x⁷ * ln(x).
A: f'(x) = 7x⁶ ln x + x⁷(1/x) = x⁶(1 + 7 ln x).
Q: Differentiate y = x² / (2x - 1).
A: y' = [2x(2x - 1) - x²(2)] / (2x - 1)² = (2x² - 2x) / (2x - 1)².
Q: Differentiate f(x) = ln(x) / (1 + x).
A: f'(x) = [(1/x)(1 + x) - ln x] / (1 + x)² = (1 + x - x ln x) / [x(1 + x)²].
Q: Differentiate y = (1 + x - x²) * e^x.
A: y' = (1 - 2x)e^x + (1 + x - x²)e^x = e^x(2 - x - x²).
Q: Find the tangent line to f(x) = (x - 8)/(3x - 4) at x = 2.
A: f(2) = -3. f'(x) = 20/(3x - 4)². f'(2) = 5. Tangent: y = 5x - 13.
Q: Find the x-values where the tangent is horizontal on f(x) = (2x - 3)(x² - 6).
A: f'(x) = 6x² - 6x - 12. Set to 0: x² - x - 2 = 0, (x - 2)(x + 1) = 0. x = 2 and x = -1.
Q: S(t) = 150t/(t + 3). Find S(12), S'(12), and estimate S(13).
A: S'(t) = 450/(t + 3)². S(12) = 120 ($120,000). S'(12) = 2 ($2,000). S(13) is approximately 122 ($122,000).
Q: P(x) = 100/x - 0.5x² + 2x - 100. Find P'(5) and interpret.
A: P'(x) = -100/x² - x + 2. P'(5) = -4 - 5 + 2 = -7. The profit is decreasing at a rate of $7 per dollar increase in price. Profit is decreasing at this price because the rate of change is negative.
The product and quotient rules build on every earlier derivative rule in the course. The marginal analysis pattern from the profit/cost/revenue section reappears here with rational functions (sales over time, subscribers over time). These problems also test the tangent-line technique from the ln/e^x section, now applied to quotients.
Looking ahead, these rules are essential for optimisation (finding maximum profit, minimum cost) and for working with more complex models in later chapters.
product rule, quotient rule, horizontal tangent, tangent line, slope-intercept form, marginal analysis, sales function, subscriber growth, derivative simplification, factoring derivatives, MAC 2233, business calculus, Exam 2, FIU, rational functions, applied derivatives, rate of change, profit as function of price