Math · Calculus I · Worked example

Differentiate (3x + 1)² with the chain rule

Differentiate y = (3x + 1)² with the chain rule, then check by expanding.

y = {(3 x + 1)}^{2}

Name the inside and the outside

Work out (3x + 1)² for one value of x: you find 3x + 1 first, then square it. So the inside is u = 3x + 1 and the outside squares it, y = u². The inner rate is du/dx = 3 and the outer rate is dy/du = 2u.

u = 3 x + 1, \frac{d u}{d x} = 3, y = u^{2}, \frac{d y}{d u} = 2 u

\begin{array}{c} u = 3 x + 1, \frac{d u}{d x} = 3 \\ y = u^{2}, \frac{d y}{d u} = 2 u \end{array}

Multiply the rates

The chain rule multiplies the two rates: dy/dx = (dy/du)(du/dx) = 2u·3. Each unit of change in x makes 3 units of change in u, and each unit of change in u makes 2u units of change in y.

\frac{d y}{d x} = 2 u \cdot 3

Put the inside back

Replace u with 3x + 1. The outer rate is evaluated at the inside, u = 3x + 1, not at x.

\frac{d y}{d x} = 2 (3 x + 1) (3)

Simplify

Multiply the numbers first, 2·3 = 6, so dy/dx = 6(3x + 1) = 18x + 6.

\frac{d y}{d x} = 18 x + 6

Check by expanding

(3x + 1)² = 9x² + 6x + 1, and the power rule gives 18x + 6. The two routes agree. For higher powers, such as (3x + 1)⁵⁰, expanding is impractical and the chain rule is the only sensible route.

\frac{d}{d x} (9 x^{2} + 6 x + 1) = 18 x + 6

Result

dy/dx = 6(3x + 1) = 18x + 6.

\frac{d y}{d x} = 18 x + 6

Your turn

Differentiate f(x) = (2x − 3)³.

Show the answer and explanation

f′(x) = 6(2x − 3)².

The outer function cubes, so its derivative is 3u² at u = 2x − 3. Multiply by the inner derivative 2: f′(x) = 3(2x − 3)²·2 = 6(2x − 3)².

\begin{array}{l} f (x) = {(2 x - 3)}^{3} \\ f^{'} (x) = 3 {(2 x - 3)}^{2} (2) \\ f^{'} (x) = 6 {(2 x - 3)}^{2} \end{array}

Keep exploring

Open Chain Rule Composition at x = 1: the inner rate 3 times the outer rate 8 gives the combined rate 24, which is 18(1) + 6.

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