Product, Quotient & Chain Rules

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Skill: derivative-rules

Product, Quotient & Chain Rules

The three rules that differentiate almost everything: the product rule for multiplied chunks, the quotient rule for fractions, and the chain rule for functions inside functions. Master these and every derivative becomes bookkeeping.

1 The product rule

When two chunks of x are multiplied, the derivative is not the product of the derivatives. Each chunk takes a turn being differentiated while the other watches.

Product rule
(fg)′ = f′g + fg′

Differentiate the first, leave the second — plus leave the first, differentiate the second.

Example Differentiate f(x) = (2x + 1)(3x − 4)
  1. Name the chunks: f = 2x + 1, g = 3x − 4.
  2. Their derivatives: f′ = 2, g′ = 3.
  3. Product rule: f′(x) = 2(3x − 4) + 3(2x + 1).
  4. Expand: 6x − 8 + 6x + 3 = 12x − 5.

Check by expanding first: (2x+1)(3x−4) = 6x2 − 5x − 4, whose derivative is 12x − 5. Same answer — the rule just skips the expansion step.

2 The quotient rule

For a fraction of two chunks, the order in the numerator matters. Memorize it as: low d-high minus high d-low, over low squared.

Quotient rule
fg′ = g·f′ − f·g′g2

“Low d-high minus high d-low, over low squared.” The minus sign is where everyone slips — say it out loud as you write.

Example Differentiate f(x) = (3x + 2)/(x − 1)
  1. High = 3x + 2 (d-high = 3); low = x − 1 (d-low = 1).
  2. Numerator: (x − 1)(3) − (3x + 2)(1) = 3x − 3 − 3x − 2 = −5.
  3. Denominator: (x − 1)2.

Answer: f′(x) = −5/(x − 1)2. Note the derivative is never zero here — the original function has no horizontal tangents.

3 The chain rule

When one function sits inside another, peel from the outside in: derivative of the outer (with the inner left alone) times the derivative of the inner.

Chain rule
d/dx [ F(u(x)) ] = F′(u) · u′(x)

Two multiplications, always: outer′ × inner′. Forgetting the second one is the most common calculus mistake there is.

inner u(x) e.g. 4x² − 3 outer F(u) e.g. u⁵ x flows in → derivatives multiply back: F′(u) × u′(x) Think of it as a two-stage machine.
The chain rule: x flows forward through inner then outer; derivatives multiply on the way back.
Example Differentiate f(x) = (4x2 − 3)5
  1. Outer: u5 → outer′ is 5u4. Inner: u = 4x2 − 3 → inner′ is 8x.
  2. Multiply: f′(x) = 5(4x2 − 3)4 · 8x = 40x(4x2 − 3)4.
Example Nested chains: differentiate f(x) = sin2(3x)
  1. Write it as (sin(3x))2 — three layers: power, sine, 3x.
  2. Peel: 2·sin(3x) × cos(3x) × 3.
  3. Answer: f′(x) = 6 sin(3x) cos(3x) (which equals 3 sin(6x)).

4 Common mistakes

Three traps, each with the wrong version and the fix.

1. Forgetting the inner derivative (chain rule)

Wrong: d/dx[(3x + 1)4] = 4(3x + 1)3.    Right: multiply by the inner derivative too: 4(3x + 1)3 · 3 = 12(3x + 1)3.

2. Quotient numerator in the wrong order

Wrong: ((3x+2)(1) − (x−1)(3))/(x−1)2 = +5/(x−1)2.    Right: low d-high minus high d-low: ((x−1)(3) − (3x+2)(1))/(x−1)2 = −5/(x−1)2.

3. “Derivative of a product = product of derivatives”

Wrong: d/dx[x2 · sin x] = 2x · cos x.    Right: (fg)′ = f′g + fg′ = 2x sin x + x2 cos x. There is no shortcut.

Key vocabulary

  • Inner / outer function — in F(u(x)), u is the inner function and F is the outer.
  • Composition — a function applied to the output of another, written F ∘ u.
  • Product rule — (fg)′ = f′g + fg′.
  • Quotient rule — (f/g)′ = (gf′ − fg′)/g2; undefined where g = 0.
  • Chain rule — d/dx[F(u(x))] = F′(u) · u′(x).

5 Quick checks

Cover the answers, decide, then reveal.

1. True or false: d/dx[x2 · sin x] = 2x · cos x.
False. The product rule gives 2x sin x + x2 cos x. Differentiating each factor separately and multiplying is never correct.
2. What is d/dx[(5x − 2)3]?
15(5x − 2)2. Outer derivative 3(5x−2)2 times inner derivative 5.
3. For f = g/h, is f′ = (g′h − gh′)/h2 right?
No — the order is flipped. It must be low d-high minus high d-low: (h·g′ − g·h′)/h2.
Practice now Try the derivative stepper Open the chain-rule lab
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