Lesson 6: Power, Product, and Quotient RulesAP

Computing derivatives from the limit definition every time would take forever; CED Unit 2 (topics 2.5–2.10) replaces it with rules. This lesson covers the working toolkit: the power rule, sum and constant-multiple rules, the derivatives of the standard functions (sin, cos, e^x, ln x), and the product and quotient rules — including the table problems, where the exam gives you values of f, g, f' and g' and asks for the derivative of a combination. Those problems are pure rule-application and are among the most reliably earned points on the test.

The basic rules

• Constant: d/dx (c) = 0
• Power rule: d/dx (x^n) = n·x^(n−1) for any real n — including fractions and negatives: d/dx (sqrt(x)) = (1/2)x^(−1/2) = 1/(2 sqrt(x)); d/dx (1/x^2) = −2/x^3.
• Constant multiple: d/dx (c·f) = c·f'
• Sum/difference: d/dx (f ± g) = f' ± g'
• Exponential: d/dx (e^x) = e^x; more generally d/dx (a^x) = a^x · ln a
• Logarithm: d/dx (ln x) = 1/x (for x > 0); d/dx (ln |x|) = 1/x for x ≠ 0
• Trig: d/dx (sin x) = cos x; d/dx (cos x) = −sin x; d/dx (tan x) = sec^2 x (from the quotient rule)

The power rule for integer n was proved by expanding (x + h)^n in Lesson 5's framework; the exam expects fluent use, not reproof — but it does expect you to rewrite before you differentiate: turn roots into fractional powers and reciprocals into negative powers first.

Worked Example 1: rewriting, then differentiating

Problem. Find f'(x) for f(x) = 4x^3 − (5/x^2) + 6 sqrt(x) − 7, and evaluate f'(1).

Rewrite:  f(x) = 4x^3 − 5x^(−2) + 6x^(1/2) − 7
Differentiate term by term:
  f'(x) = 12x^2 − 5(−2)x^(−3) + 6(1/2)x^(−1/2) − 0
        = 12x^2 + 10/x^3 + 3/sqrt(x)
f'(1) = 12 + 10 + 3 = 25

Answer: f'(x) = 12x^2 + 10/x^3 + 3/sqrt(x); f'(1) = 25. The two classic slips are both visible here: the negative exponent term *adds* (two negatives), and the constant −7 vanishes.

The product and quotient rules

Product rule: (f·g)' = f'·g + f·g' — "derivative of the first times the second, plus the first times the derivative of the second." It comes from the limit definition applied to f(x+h)g(x+h) − f(x)g(x); the derivative of a product is not the product of the derivatives.

Quotient rule: (f/g)' = (f'·g − f·g')/g^2. The order in the numerator matters and the whole numerator is divided by g^2.

Worked Example 2: product rule with a tangent line

Problem. Let h(x) = x^2 · e^x. Find h'(1) and the tangent line to h at x = 1.

h'(x) = 2x·e^x + x^2·e^x = e^x (x^2 + 2x)
h(1) = e,  h'(1) = e(1 + 2) = 3e
Tangent at (1, e):  y − e = 3e(x − 1)

Answer: h'(1) = 3e ≈ 8.155; tangent y = e + 3e(x − 1). Leaving answers in exact form (3e, not 8.155) is the exam's preference unless a decimal is requested.

Worked Example 3: the table problem (graphical/numerical, College Board style)

Problem. The table gives values of differentiable functions f and g and their derivatives at x = 2. Let H(x) = f(x)·g(x) and K(x) = f(x)/g(x). Find H'(2) and K'(2).

  x = 2:   f = 3    g = 5    f' = −1    g' = 4

H'(2) = f'(2)g(2) + f(2)g'(2) = (−1)(5) + (3)(4) = −5 + 12 = 7
K'(2) = [f'(2)g(2) − f(2)g'(2)] / g(2)^2 = [(−1)(5) − (3)(4)] / 25 = −17/25

Answer: H'(2) = 7; K'(2) = −17/25. No formulas for f and g are needed or given — this is the exam testing whether the rules are understood as *operations on values*. Keep the four table entries straight: the most common error is swapping an f' value with a g value.

Real-world application: marginal analysis in business

A firm sells x thousand units at a price that depends on volume: p(x) = 60 − 2x dollars. Revenue is R(x) = x · p(x) (a product), and the marginal revenue R'(x) is the approximate revenue from one more unit. R(x) = 60x − 2x^2, so R'(x) = 60 − 4x; at x = 10, R'(10) = 20 dollars per unit — selling the 11th thousand units adds about 20 thousand dollars, not the 40 thousand the current price naively suggests, because the price had to drop on *all* units. When R' changes sign (here at x = 15), revenue is maximised — the calculus behind pricing decisions, and a preview of Lesson 10's optimisation.

Common mistakes

• (fg)' = f'g'. The single most-tested misconception in Unit 2. If a problem hands you a product, the product rule is compulsory.
• Quotient rule order and square: writing (f'g − fg')/g instead of /g^2, or reversing to (fg' − f'g). "Low-dee-high minus high-dee-low, over low squared" — pick a memory hook and keep it.
• Differentiating e^x with the power rule (getting x·e^(x−1)) or applying the power rule to 2^x (getting x·2^(x−1) instead of 2^x ln 2). The variable's location decides the rule: variable in the base → power rule; variable in the exponent → exponential rule.
• Forgetting to rewrite 1/x^3 as x^(−3) and differentiating "the fraction" by instinct.

AP Exam Experience

1. (No calculator.) If f(x) = x^5 − 3x^2 + 7, then f'(x) =

• (A) 5x^4 − 6x
• (B) 5x^4 − 6x + 7
• (C) x^4 − 6x + 7
• (D) 5x^4 − 3x

Answer: (A). Term-by-term power rule; the constant 7 differentiates to 0. (D) is the trap for treating −3x^2 as −3x^1 after differentiating.

2. (No calculator.) d/dx (x^3 · cos x) =

• (A) 3x^2 · (−sin x)
• (B) 3x^2 cos x − x^3 sin x
• (C) 3x^2 cos x + x^3 sin x
• (D) −3x^2 sin x

Answer: (B). Product rule: derivative of x^3 times cos x, plus x^3 times derivative of cos x (which is −sin x). (A) is the f'g' fallacy with a sign error added.

3. The table shows f(1) = 2, g(1) = 1, f'(1) = 4, g'(1) = 3. If H(x) = f(x)/(g(x) + 1), then H'(1) =

• (A) 1/2
• (B) 2
• (C) 5/4
• (D) −1/2

Answer: (A). Quotient rule, noting the denominator g(x) + 1 has derivative g'(x): H'(1) = [f'(1)(g(1) + 1) − f(1)·g'(1)]/(g(1) + 1)^2 = [4·2 − 2·3]/(2)^2 = (8 − 6)/4 = 1/2. The two classic errors this item is built on: forgetting the +1 when evaluating the denominator, and using g(1) where g'(1) belongs.

Free-response question. Let f and g be differentiable functions with the following values: f(2) = 5, f'(2) = −3, g(2) = 1, g'(2) = 4. Let h(x) = f(x)·g(x) and let k(x) = (f(x))/(x + g(x)).

(a) Find h'(2). (b) Find k'(2). (c) Write the equation of the tangent line to the graph of h at x = 2.

Scoring guideline.

• (a) 1 point. h'(2) = f'(2)g(2) + f(2)g'(2) = (−3)(1) + (5)(4) = 17. The point requires the product-rule structure with table values substituted — an unsupported 17 earns nothing.
• (b) 2 points. 1 point for the quotient-rule setup with denominator derivative 1 + g'(x) (the derivative of x + g(x)); 1 point for the value: k'(2) = [f'(2)(2 + g(2)) − f(2)(1 + g'(2))]/(2 + g(2))^2 = [(−3)(3) − (5)(5)]/9 = (−9 − 25)/9 = −34/9.
• (c) 1 point. h(2) = f(2)g(2) = 5, slope h'(2) = 17, so the tangent is y − 5 = 17(x − 2). Using the derivative value from (a) even if (a) was wrong still earns this point (the exam's "consistent answer" policy), provided the point (2, 5) is correct.

Additional worked practice

Problem. Let f(x) = (x^2 + 1)/(x − 3). Find f'(4) and all x where the tangent to f is horizontal.

Quotient rule: f'(x) = [2x(x − 3) − (x^2 + 1)(1)]/(x − 3)^2 = [2x^2 − 6x − x^2 − 1]/(x − 3)^2 = (x^2 − 6x − 1)/(x − 3)^2
f'(4) = (16 − 24 − 1)/1 = −9.
Horizontal tangents: numerator zero → x^2 − 6x − 1 = 0 → x = 3 ± sqrt(10).
  (x = 3 itself is excluded — f is not defined there; a vertical asymptote is not a tangent point.)

Answers: f'(4) = −9; horizontal tangents at x = 3 ± sqrt(10) ≈ −0.16 and 6.16. Note the division of labour: the *numerator* of a quotient-rule derivative finds horizontal tangents, the *denominator* (squared, hence nonnegative) never affects the sign — f' here is negative between the two roots and positive outside, information Lesson 12 will cash in.

Extension: higher-order derivatives and the derivative of tan x

Derivatives can be differentiated. The second derivative f'' measures the rate of change of the rate of change — acceleration when f is position. If s(t) = t^3 − 6t^2 + 9t is a particle's position (metres), then velocity v(t) = s'(t) = 3t^2 − 12t + 9 and acceleration a(t) = s''(t) = 6t − 12. At t = 2: v = 12 − 24 + 9 = −3 m/s (moving in the negative direction) and a = 0 — the instant the particle stops accelerating and begins speeding up in the negative direction... precisely, a changes sign at t = 2, so the velocity stops decreasing and starts increasing there. Unit 5 will read whole graphs from f''; this is the vocabulary.

One more rule completes the trig set, derived rather than memorised: tan x = sin x / cos x, so by the quotient rule, d/dx tan x = (cos x · cos x − sin x · (−sin x))/cos^2 x = (cos^2 x + sin^2 x)/cos^2 x = sec^2 x. The Pythagorean identity does the cleaning. The same quotient-rule derivation regenerates any forgotten trig derivative on demand — cot, sec and csc are BC-adjacent, but the *method* is the portable asset. Finally, the general exponential: a^x = e^(x ln a), so by the chain rule d/dx a^x = e^(x ln a) · ln a = a^x ln a — which is why d/dx 2^x = 2^x ln 2, and why e, with ln e = 1, is the unique base whose derivative needs no correction factor.