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Calculus Honors · AP-aligned study guide

2.3 · Product, quotient, and chain rules

Recognize the structure of a function before choosing a rule, including products containing a composition.

AP 2.9–2.10; 3.1; 3.5 · Reading: Larson & Edwards §§2.3–2.4

Learning objectives

  • Select and combine product, quotient, and chain rules.
  • Identify the inner derivative in a composition.
  • Check derivative expressions at specified inputs.

A product needs two terms

The derivative of a product is not the product of the derivatives. Each factor contributes a term: keep one unchanged while differentiating the other, then add.

For x2ex\displaystyle x^2e^x, the derivative is 2xex+x2ex\displaystyle 2xe^x+x^2e^x. Factor afterward if useful; expanding first is optional.

(uv)′=u′v+uv′(uv)'=u' v+uv'

A quotient needs a squared denominator

For u/v\displaystyle u/v, with v≠0\displaystyle v\ne0, the derivative numerator is denominator times derivative of numerator minus numerator times derivative of denominator. Preserve the order to avoid a sign reversal.

For (x+1)/(x−1)\displaystyle (x+1)/(x-1), the numerator of the derivative is (x−1)−(x+1)=−2\displaystyle (x-1)-(x+1)=-2, giving −2/(x−1)2\displaystyle -2/(x-1)^2. The excluded input remains excluded.

(uv)′=u′v−uv′v2\left(\frac{u}{v}\right)'=\frac{u' v-uv'}{v^2}

A composition needs the inner rate

For f(g(x))\displaystyle f(g(x)), first differentiate the outer function while retaining the inner input, then multiply by g′(x)\displaystyle g'(x). For (3x+1)4\displaystyle (3x+1)^4, this gives 4(3x+1)3⋅3\displaystyle 4(3x+1)^3\cdot3.

A product may contain a composition. For xsin⁡(x2)\displaystyle x\sin(x^2), product differentiation gives sin⁡(x2)+xcos⁡(x2)⋅2x\displaystyle \sin(x^2)+x\cos(x^2)\cdot2x. Identify the structure at each level rather than choosing only one rule.

ddxf(g(x))=f′(g(x))g′(x)\frac{d}{dx}f(g(x))=f'(g(x))g'(x)

Worked examples

Product with a composition

Differentiate x2sin⁡(3x)\displaystyle x^2\sin(3x).

  1. Use the product rule: 2xsin⁡(3x)+x2ddxsin⁡(3x)\displaystyle 2x\sin(3x)+x^2\frac{d}{dx}\sin(3x).
  2. The chain rule gives ddxsin⁡(3x)=3cos⁡(3x)\displaystyle \frac{d}{dx}\sin(3x)=3\cos(3x).
  3. The derivative is 2xsin⁡(3x)+3x2cos⁡(3x)\displaystyle 2x\sin(3x)+3x^2\cos(3x).

A quotient at one point

Find f′(0)\displaystyle f'(0) if f(x)=exx+1\displaystyle f(x)=\frac{e^x}{x+1}.

  1. The quotient rule gives f′(x)=ex(x+1)−ex(x+1)2\displaystyle f'(x)=\frac{e^x(x+1)-e^x}{(x+1)^2}.
  2. Simplify the numerator to xex\displaystyle xe^x.
  3. At x=0\displaystyle x=0, the result is 0\displaystyle 0.

Common mistakes

  • Differentiating a product by multiplying the two derivatives.
  • Reversing the subtraction in the quotient rule.
  • Forgetting the inner derivative.

Try it yourself

No calculator required.

1. Differentiate f(x)=(2x2+1)3\displaystyle f(x)=(2x^2+1)^3.

Show worked solution

The outer derivative is 3(2x2+1)2\displaystyle 3(2x^2+1)^2 and the inner derivative is 4x\displaystyle 4x, giving f′(x)=12x(2x2+1)2\displaystyle f'(x)=12x(2x^2+1)^2.

2. Differentiate g(x)=x2e2x\displaystyle g(x)=x^2e^{2x}.

Show worked solution

Use product and chain rules: g′(x)=2xe2x+2x2e2x\displaystyle g'(x)=2xe^{2x}+2x^2e^{2x}.

3. Find h′(2)\displaystyle h'(2) for h(x)=x2x+1\displaystyle h(x)=\frac{x^2}{x+1}.

Show worked solution

The quotient rule gives h′(x)=2x(x+1)−x2(x+1)2\displaystyle h'(x)=\frac{2x(x+1)-x^2}{(x+1)^2}. Substituting 2\displaystyle 2 gives 89\displaystyle \frac{8}{9}.

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