Math · Calculus I · Worked example

Combine the chain rule and the product rule

Differentiate y = x²e^(3x).

y = x^{2} e^{3 x}

Find the top-level structure

Work out x²e^(3x) for x = 1: you find x² = 1 and e^(3·1) = e³ separately, then multiply them. The last operation is a multiplication, so the function is a product, and the product rule comes first.

Write the product rule

The derivative of a product is the first factor’s derivative times the second factor, plus the first factor times the second factor’s derivative. Here the first factor is x² and the second is e^(3x).

{(f g)}^{'} = f^{'} g + f g^{'}

Differentiate the first factor

x² needs only the power rule.

\frac{d}{d x} x^{2} = 2 x

Differentiate the second factor with the chain rule

e^(3x) is a composition: the inside is 3x and the outside is eᵘ. The outside gives e^(3x), copied unchanged, and the inside gives 3.

\frac{d}{d x} e^{3 x} = e^{3 x} \cdot 3 = 3 e^{3 x}

Put the product rule together

Substitute both derivatives into the product rule.

\frac{d y}{d x} = 2 x \cdot e^{3 x} + x^{2} \cdot 3 e^{3 x}

Factor the answer

Both terms contain x and e^(3x), so factor them out. The factored form shows where the slope is zero: e^(3x) is never 0, so dy/dx = 0 only at x = 0 and at x = −2/3.

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

Result

dy/dx = 2xe^(3x) + 3x²e^(3x) = xe^(3x)(2 + 3x).

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

Your turn

Differentiate y = x(2x + 1)⁴.

Show the answer and explanation

dy/dx = (2x + 1)³(10x + 1).

The top level is the product of x and (2x + 1)⁴. The product rule gives 1·(2x + 1)⁴ + x·[(2x + 1)⁴]′, and the chain rule gives [(2x + 1)⁴]′ = 4(2x + 1)³·2 = 8(2x + 1)³. So dy/dx = (2x + 1)⁴ + 8x(2x + 1)³. Factoring out (2x + 1)³ leaves (2x + 1) + 8x = 10x + 1.

\begin{array}{l} y = x {(2 x + 1)}^{4} \\ \frac{d y}{d x} = {(2 x + 1)}^{4} + 8 x {(2 x + 1)}^{3} \\ \frac{d y}{d x} = {(2 x + 1)}^{3} (10 x + 1) \end{array}

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