Math · Calculus I · Worked example

Differentiate x²(x + 1) with the product rule

Differentiate f(x) = x²(x + 1) with the product rule, then check the result by expanding first.

f (x) = x^{2} (x + 1)

Name the two factors

Write f(x) = u(x)v(x) with u = x² and v = x + 1. Each factor has a simple derivative: u′ = 2x and v′ = 1.

u = x^{2}, u^{'} = 2 x, v = x + 1, v^{'} = 1

\begin{array}{c} u = x^{2}, u^{'} = 2 x \\ v = x + 1, v^{'} = 1 \end{array}

Apply the product rule

Differentiate the first factor and keep the second, then keep the first factor and differentiate the second.

f^{'} (x) = 2 x (x + 1) + x^{2} (1)

Simplify

Expand and collect like terms: 2x² + 2x + x² = 3x² + 2x.

f^{'} (x) = 3 x^{2} + 2 x

Check by expanding first

Multiplying out gives f(x) = x³ + x², and the power rule gives 3x² + 2x. The two methods agree.

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

Result

f′(x) = 3x² + 2x.

f^{'} (x) = 3 x^{2} + 2 x

Your turn

Differentiate g(x) = x³eˣ.

Show the answer and explanation

g′(x) = 3x²eˣ + x³eˣ = x²eˣ(x + 3).

The first factor x³ has derivative 3x² and the second factor eˣ is its own derivative. The product rule gives 3x²·eˣ + x³·eˣ, and factoring out x²eˣ gives x²eˣ(x + 3).

\begin{array}{l} g (x) = x^{3} e^{x} \\ g^{'} (x) = 3 x^{2} e^{x} + x^{3} e^{x} \end{array}

Keep exploring

Open Product Rule Geometry at x = 1 and shrink h: the two edge strips give the derivative 5, and the corner vanishes.

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