Lesson 8: Product & Quotient Rules

(a) Opening Question

You already know how to differentiate x² and x³. The power rule gives 2x and 3x². So here is a tempting shortcut. Consider:

y = x² · x³

A student reasons: "The derivative of a product is the product of the derivatives, so y' = (2x)(3x²) = 6x³."

Is that correct? Check it a different way.

Notice x² · x³ = x⁵. By the power rule, the derivative of x⁵ is 5x⁴.

But the shortcut gave 6x³. These are not equal — not even close. So the derivative of a product is NOT the product of the derivatives. Something more careful is needed.

This single observation is the reason the Product Rule exists, and it is the most important idea in this lesson. Hold onto the contradiction 6x³ ≠ 5x⁴; by the end of section (b) you'll see exactly how the real rule fixes it.

(b) Core Concepts

Why a simple product shortcut fails

The opening question proves it numerically: d/dx[x²·x³] cannot be (2x)(3x²). The correct rule accounts for the fact that both factors are changing at once. When you nudge x, the first factor changes a little, the second factor changes a little, and the product responds to both changes.

The Product Rule

Product Rule. If f and g are differentiable, then

`

d/dx[ f · g ] = f' · g + f · g'

`

In words: (derivative of the first)(second) + (first)(derivative of the second).

Each term keeps one factor undifferentiated. There are two terms because there are two factors, each getting its turn to be differentiated.

Check it against the opening question. With f = x² and g = x³:

f' · g + f · g' = (2x)(x³) + (x²)(3x²) = 2x⁴ + 3x⁴ = 5x⁴ ✓

That matches the power rule on x⁵. The rule works.

Fully worked Product-Rule example. Differentiate y = (3x² + 1)(2x − 5).

Identify the factors:

f = 3x² + 1      f' = 6x
g = 2x − 5       g' = 2

Apply the rule:

y' = f'·g + f·g'
   = (6x)(2x − 5) + (3x² + 1)(2)
   = (12x² − 30x) + (6x² + 2)
   = 18x² − 30x + 2

Verification. Expand first: (3x² + 1)(2x − 5) = 6x³ − 15x² + 2x − 5, whose derivative is 18x² − 30x + 2. ✓ The two routes agree.

The Quotient Rule

Quotient Rule. If f and g are differentiable and g ≠ 0, then

`

f' · g − f · g'

d/dx[ f/g ] = ───────────────

g²

`

Two things make the Quotient Rule the most error-prone rule in Unit 2:

1. It is a subtraction, not addition — opposite of the Product Rule.
2. Order matters. The term f'·g comes first. Swap the two terms and you flip the sign of the whole answer, which is wrong.

Memory device for the order. Many students use the chant

"Lo · D-Hi − Hi · D-Lo, all over Lo-Lo."

Here Hi is the top (numerator), Lo is the bottom (denominator), and D means "derivative of." So:

        Lo·D(Hi) − Hi·D(Lo)
d/dx[Hi/Lo] = ─────────────────────
                   (Lo)²

The bottom function Lo (which is g) leads off the numerator, multiplied by the derivative of the top. The single most common mistake is starting with the top — so anchor on "Lo D-Hi first."

Fully worked Quotient-Rule example. Differentiate y = (2x + 1)/(x² + 3).

Identify top and bottom:

f (Hi) = 2x + 1      f' = 2
g (Lo) = x² + 3      g' = 2x

Apply the rule — Lo·D-Hi first:

     f'·g − f·g'      (2)(x² + 3) − (2x + 1)(2x)
y' = ───────────  =  ──────────────────────────────
        g²                    (x² + 3)²

     (2x² + 6) − (4x² + 2x)        −2x² − 2x + 6
   = ───────────────────────  =  ─────────────────
          (x² + 3)²                 (x² + 3)²

Notice we expand and simplify the numerator only. Leave the denominator as (x² + 3)² — on the AP exam you almost never multiply it out.

Table-of-values problems

A favorite AP format gives you values of f, g, f', and g' at specific points and asks for the derivative of a product or quotient there. You never need a formula — just plug numbers into the rule.

Worked table example. Suppose at x = 2 we know:

Let h(x) = f(x)·g(x). Find h'(2).

h'(2) = f'(2)·g(2) + f(2)·g'(2)
      = (−1)(3) + (5)(4)
      = −3 + 20 = 17

Let k(x) = f(x)/g(x). Find k'(2).

        f'(2)·g(2) − f(2)·g'(2)      (−1)(3) − (5)(4)      −3 − 20      −23
k'(2) = ─────────────────────────  = ───────────────────  = ─────────  = ────
              [g(2)]²                       3²                  9          9

When NOT to use the Quotient Rule

The Quotient Rule is mandatory only when the variable sits in the denominator in a way you can't easily separate. Often it's faster — and less error-prone — to simplify first:

• y = (x³ + 2x)/x → rewrite as y = x² + 2, so y' = 2x. No Quotient Rule needed.
• y = (5x − 2)/x² → rewrite as y = 5x⁻¹ − 2x⁻², so y' = −5x⁻² + 4x⁻³.

If the denominator is a single power of x (or a single monomial), split the fraction and use the power rule. Save the Quotient Rule for genuine quotients like (2x+1)/(x²+3), where the denominator can't be split off.

Calculator note. This lesson is primarily non-calculator — the AP exam expects you to apply these rules by hand. You may use nDeriv only as a check: e.g. TI-84: MATH → 8:nDeriv( → nDeriv((2X+1)/(X²+3), X, 1) returns about 0.125, which confirms our (2x+1)/(x²+3) derivative gives (−2−2+6)/16 = 2/16 = 0.125 at x = 1. Use it to catch errors, never to replace the algebra.

(c) Worked Examples

Example 1 — Product Rule (foundation) · [NO CALC]

Problem. Differentiate y = (x² + 2x)(x³ − 1).

Strategy. Two genuine factors → Product Rule. Identify f, g, and their derivatives.

Solution.

f = x² + 2x      f' = 2x + 2
g = x³ − 1       g' = 3x²

y' = f'·g + f·g'
   = (2x + 2)(x³ − 1) + (x² + 2x)(3x²)
   = (2x⁴ + 2x³ − 2x − 2) + (3x⁴ + 6x³)
   = 5x⁴ + 8x³ − 2x − 2

Justification / check. Expanding first gives x⁵ + 2x⁴ − x² − 2x; its derivative is 5x⁴ + 8x³ − 2x − 2. ✓ Agreement confirms the work.

Example 2 — Quotient Rule evaluated at a point (procedural) · [NO CALC]

Problem. Let y = (x² − 1)/(x + 2). Find y'(1).

Strategy. Genuine quotient (denominator can't be split cleanly) → Quotient Rule, Lo·D-Hi first, then substitute x = 1.

Solution.

f (Hi) = x² − 1     f' = 2x
g (Lo) = x + 2      g' = 1

     f'·g − f·g'      (2x)(x + 2) − (x² − 1)(1)
y' = ───────────  =  ──────────────────────────────
        g²                    (x + 2)²

Substitute x = 1 (no need to fully simplify symbolically first):

       (2)(3) − (0)(1)      6 − 0      6      2
y'(1) = ─────────────────  = ───────  = ───  = ───
            (3)²               9         9      3

Answer. y'(1) = 2/3.

Check. Simplified numerator is (2x² + 4x) − (x² − 1) = x² + 4x + 1, so y' = (x² + 4x + 1)/(x+2)²; at x = 1 that's (1+4+1)/9 = 6/9 = 2/3. ✓

Example 3 — Combined product + quotient (AP level) · [NO CALC]

Problem. Differentiate y = x(x² + 1)/(x − 1).

Strategy. A product x(x² + 1) sits on top of a quotient. Treat the numerator as a single function f and differentiate it with the Product Rule (or just expand it), then apply the Quotient Rule.

Solution. First combine the top: x(x² + 1) = x³ + x. Now it's a clean quotient.

f (Hi) = x³ + x      f' = 3x² + 1
g (Lo) = x − 1       g' = 1

     (3x² + 1)(x − 1) − (x³ + x)(1)
y' = ────────────────────────────────
                 (x − 1)²

Expand the numerator:

(3x² + 1)(x − 1) = 3x³ − 3x² + x − 1
subtract (x³ + x):  3x³ − 3x² + x − 1 − x³ − x = 2x³ − 3x² − 1

     2x³ − 3x² − 1
y' = ───────────────
        (x − 1)²

Justification. Expanding the numerator-product before differentiating turned a product-inside-a-quotient into a single Quotient Rule application — a standard simplify-first move that avoids nested rules. (sympy confirms y' = (2x³ − 3x² − 1)/(x − 1)².) ✓

Example 4 — Table of values at a point (AP MCQ format) · [NO CALC]

Problem. The table gives values of f, g, and their derivatives at x = 4.

Let P(x) = f(x)·g(x) and Q(x) = g(x)/f(x). Find P'(4) and Q'(4).

Strategy. Plug the table values directly into the Product and Quotient Rules. For Q, the top is g and the bottom is f — watch the order.

Solution.

P'(4) = f'(4)·g(4) + f(4)·g'(4) = (3)(5) + (6)(2) = 15 + 12 = 27

        g'(4)·f(4) − g(4)·f'(4)      (2)(6) − (5)(3)      12 − 15      −3      1
Q'(4) = ─────────────────────────  = ────────────────────  = ─────────  = ────  = − ───
              [f(4)]²                        6²                 36       36       12

Answer. P'(4) = 27, Q'(4) = −1/12.

Justification. Because Q = g/f, the bottom function in the Quotient Rule is f, so the denominator is [f(4)]² = 36 and the numerator leads with g'·f (Lo·D-Hi, where here "Hi" = g). Reversing top and bottom would give the wrong sign and wrong denominator.

(d) Common Mistakes

Thinking d/dx[f·g] = f'·g'. This is the trap from the Opening Question. Students multiply the two derivatives. ✗ The correct Product Rule keeps one factor undifferentiated in each term: f'·g + f·g'. Quick gut-check: on x²·x³ the shortcut gives 6x³ but the truth is 5x⁴.

Quotient Rule order/sign — starting with the top. Writing (f·g' − f'·g)/g² (Hi-D-Lo first) flips the sign of the entire answer. ✗ Anchor on "Lo·D-Hi first": the numerator must be f'·g − f·g'. The first term differentiates the top and keeps the bottom.

Forgetting to square the denominator. Students write the numerator correctly but divide by g instead of g². ✗ The denominator is always g² ((Lo)²). Even if the original denominator was x + 2, the derivative's denominator is (x + 2)².

Differentiating the squared denominator. After writing g², some students "helpfully" apply the chain rule to it. ✗ The g² in the Quotient Rule formula is part of the formula, not something to differentiate. Leave it as is and simplify only the numerator.

Not simplifying when simplifying is easier. Reaching for the Quotient Rule on (5x − 2)/x² when splitting into 5x⁻¹ − 2x⁻² is faster and cleaner. ✗ Always glance at the denominator first: a single monomial → split and use the power rule; a genuine binomial/expression → Quotient Rule.

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## (e) Practice Problems

Attempt before checking section (g). [NO CALC] unless marked [CALC].

Find dy/dx for y = √x·(x + 1). Write your answer as a single simplified fraction.

Find f'(x) for f(x) = x²(2x − 1). Then evaluate f'(1).

(justify) A student claims d/dx[(x − 1)/(x + 1)] = 1 because "the derivative of the top is 1 and the derivative of the bottom is 1, so 1/1 = 1." Compute the correct derivative and, in one sentence, explain the student's error.

(justify) Let g(x) = (3x² + 2)/(x − 5). A classmate writes the numerator of g' as (x − 5)(6x) − (3x² + 2)(1). Without computing the final answer, state whether the order and sign of this numerator are correct, and justify using the Quotient Rule.

The table gives values at x = 2. Let R(x) = g(x)/f(x). Find R'(2).

| x | f | f' | g | g' |

|---|---|---|---|---|

| 2 | 4 | 1 | 6 | −3 |

Differentiate y = (x² − 4)/(x + 2) two ways: (i) by the Quotient Rule, and (ii) by simplifying first. Confirm they agree.

Let P(x) = f(x)·g(x) where f(2) = 3, f'(2) = −2, g(2) = 5, g'(2) = 1. Find P'(2) and write the equation of the tangent line to P at x = 2.

Differentiate y = (x² + 1)(x − 3)/(x). (Hint: simplify the product over x before differentiating — decide whether the Quotient Rule is even needed.)

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## (f) AP Exam Focus — Free Response

> Free Response Question (No Calculator) — 9 points total

>

> The table below gives values of the differentiable functions f and g and their derivatives f' and g' at selected values of x.

>

> | x | f(x) | f'(x) | g(x) | g'(x) |

> |---|---|---|---|---|

> | 1 | 3 | 4 | 2 | 1 |

> | 2 | 5 | −1 | 1 | 3 |

> | 3 | 2 | 2 | 4 | −2 |

> | 4 | 6 | 3 | 5 | 2 |

>

> (a) (2 points) Let h(x) = f(x)·g(x). Find h'(2).

>

> (b) (2 points) Let k(x) = f(x)/g(x). Find k'(1). Show the setup that leads to your answer.

>

> (c) (3 points) Let h(x) = f(x)·g(x). Write an equation for the line tangent to the graph of h at x = 3, and use it to approximate h(3.1).

>

> (d) (2 points) Let p(x) = g(x)/f(x). A student claims that because g'(4) > 0, the function p must be increasing at x = 4. Find p'(4) and state whether the student is correct. Justify your answer.

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### MODEL SOLUTION

(a) h'(2) — 2 points

By the Product Rule, h'(x) = f'(x)·g(x) + f(x)·g'(x). At x = 2:

h'(2) = f'(2)·g(2) + f(2)·g'(2)
      = (−1)(1) + (5)(3)
      = −1 + 15 = 14

h'(2) = 14.

(b) k'(1) — 2 points

By the Quotient Rule, k'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]². At x = 1:

        f'(1)·g(1) − f(1)·g'(1)      (4)(2) − (3)(1)      8 − 3      5
k'(1) = ─────────────────────────  = ───────────────────  = ─────────  = ───
              [g(1)]²                       (2)²               4         4

k'(1) = 5/4.

(c) Tangent line to h at x = 3; approximate h(3.1) — 3 points

First the point: h(3) = f(3)·g(3) = (2)(4) = 8.

Then the slope (Product Rule at x = 3):

h'(3) = f'(3)·g(3) + f(3)·g'(3) = (2)(4) + (2)(−2) = 8 − 4 = 4

Tangent line through (3, 8) with slope 4:

y = 8 + 4(x − 3)

Approximate h(3.1):

h(3.1) ≈ 8 + 4(3.1 − 3) = 8 + 4(0.1) = 8.4

Tangent line: y = 8 + 4(x − 3); h(3.1) ≈ 8.4.

(d) p'(4) and justification — 2 points

By the Quotient Rule, with top g and bottom f:

        g'(4)·f(4) − g(4)·f'(4)      (2)(6) − (5)(3)      12 − 15      −3       1
p'(4) = ─────────────────────────  = ───────────────────  = ─────────  = ────  = − ───
              [f(4)]²                       (6)²               36        36       12

Since p'(4) = −1/12 < 0, the function p is decreasing at x = 4, so the student is incorrect. The sign of g'(4) alone does not determine whether p = g/f is increasing; the Quotient Rule shows the behavior of p depends on both functions and both derivatives, and here the term −g(4)·f'(4) = −15 outweighs g'(4)·f(4) = 12, making p'(4) negative.

SCORING COMMENTARY — where students lose points

Part (a) — 2 pts: 1 pt for correct Product-Rule setup f'(2)g(2) + f(2)g'(2); 1 pt for the answer 14. Common loss: using f'(2)·g'(2) (the "product of derivatives" error) earns 0/2.

Part (b) — 2 pts: 1 pt for the correct Quotient-Rule expression with [g(1)]² in the denominator and the terms in the right order; 1 pt for 5/4. Common loss: reversing the numerator to f·g' − f'·g gives −5/4 — wrong sign, loses the answer point. Forgetting to square g(1) (dividing by 2 instead of 4) also loses the answer point.

Part (c) — 3 pts: 1 pt for h(3) = 8 (the point); 1 pt for h'(3) = 4 (the slope, requires correct Product Rule); 1 pt for a correct tangent equation AND the approximation 8.4. Common loss: finding the slope but never writing an actual line equation, or computing h'(3) with a sign slip on f(3)·g'(3) = 2(−2) = −4. A bare answer of 8.4 with no tangent line shown typically earns only the point(s) the supporting work justifies — show the line.

Part (d) — 2 pts: 1 pt for p'(4) = −1/12 (correct order, correct [f(4)]² = 36); 1 pt for the justification that p is decreasing because p'(4) < 0, so the student is wrong. Common loss: getting −1/12 but giving a vague justification ("it's negative so no") without connecting the sign of p' to decreasing loses the reasoning point. Stating "the student is correct" contradicts your own derivative and loses the point even with −1/12 shown.

Notation reminder: Throughout, label which rule you use and keep [g]² written explicitly. AP readers reward f'(2)·g(2) + f(2)·g'(2) written out over a lone final number.