Math · Calculus I · Worked example

Differentiate x²(x + 1) with the product rule

Differentiate $f (x) = x^{2} (x + 1)$ 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 $u = x^{2}$ and $v = x + 1$ . Each factor has a simple derivative, $u^{'} = 2 x$ 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. The rule allows this because both factors have a derivative at every $x$ .

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

Simplify

Expand and collect like terms. $2 x^{2} + 2 x + x^{2} = 3 x^{2} + 2 x$ .

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

Check by expanding first

Multiplying out gives $f (x) = x^{3} + x^{2}$ , and the power rule gives $3 x^{2} + 2 x$ . The two methods agree. At $x = 2$ both give $16$ , the rate the nudge of $0.01$ suggested.

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

Result

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

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

Your turn

1. One like the example

Differentiate $f (x) = (2 x + 1) (x^{2} - 3)$ , then check by multiplying out.

Show the answer and explanation

$f^{'} (x) = 6 x^{2} + 2 x - 6$ .

The factors $2 x + 1$ and $x^{2} - 3$ have derivatives $2$ and $2 x$ . The product rule gives $2 (x^{2} - 3) + (2 x + 1) (2 x)$ $= 2 x^{2} - 6 + 4 x^{2} + 2 x$ $= 6 x^{2} + 2 x - 6$ . Multiplying out first gives $2 x^{3} + x^{2} - 6 x - 3$ , and the power rule gives the same derivative.

\begin{array}{l} f (x) = (2 x + 1) (x^{2} - 3) \\ f^{'} (x) = 2 (x^{2} - 3) + (2 x + 1) (2 x) \\ f^{'} (x) = 6 x^{2} + 2 x - 6 \end{array}

2. A variation

Differentiate $g (x) = x^{3} e^{x}$ . This product cannot be multiplied out.

Show the answer and explanation

$g^{'} (x)$ $= 3 x^{2} e^{x} + x^{3} e^{x}$ $= x^{2} e^{x} (x + 3)$ .

The first factor $x^{3}$ has derivative $3 x^{2}$ , and the second factor $e^{x}$ is its own derivative. The product rule gives $3 x^{2} \cdot e^{x} + x^{3} \cdot e^{x}$ , and factoring out $x^{2} e^{x}$ gives $x^{2} e^{x} (x + 3)$ .

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

3. Somewhere new

A garden bed is being enlarged. After $t$ weeks it is $10 + 2 t$ meters long and $4 + t$ meters wide. How fast is its area growing at $t = 3$ ?

Show the answer and explanation

$A^{'} (3) = 30 m^{2}$ per week.

The area is the product $A (t) = (10 + 2 t) (4 + t)$ . The length grows at $2 m$ per week and the width at $1 m$ per week. The product rule gives $A^{'} (t)$ $= 2 (4 + t) + (10 + 2 t) (1)$ $= 4 t + 18$ . At $t = 3$ the bed is $16 m$ by $7 m$ , and $A^{'} (3)$ $= 2 \cdot 7 + 16 \cdot 1$ $= 30 m^{2}$ per week.

\begin{array}{l} A (t) = (10 + 2 t) (4 + t) \\ A^{'} (t) = 2 (4 + t) + (10 + 2 t) (1) \\ A^{'} (t) = 4 t + 18 \\ A^{'} (3) = 30 \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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Name the two factors

Apply the product rule

Simplify

Check by expanding first

Result

Your turn

1. One like the example

f′(x)=6x2+2⁢x−6f′(x)=6x2+2x−6.

2. A variation

g′(x)g′(x) =3x2ex+x3ex=3x2ex+x3ex =x2ex(x+3)=x2ex(x+3).

3. Somewhere new

A′(3)=30 m2A′(3)=30 m2 per week.

Keep exploring

Study it, then try it

$f^{'} (x) = 6 x^{2} + 2 x - 6$ .

$g^{'} (x)$ $= 3 x^{2} e^{x} + x^{3} e^{x}$ $= x^{2} e^{x} (x + 3)$ .

$A^{'} (3) = 30 m^{2}$ per week.