Math · Calculus I · Worked example

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

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

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

Name the inside and the outside

Work out ${(3 x + 1)}^{2}$ for one value of $x$ : you find $3 x + 1$ first, then square it. The inside is $u = 3 x + 1$ and the outside squares it, $y = u^{2}$ . The inner rate is $d u / d x = 3$ and the outer rate is $d y / d u = 2 u$ .

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: $d y / d x$ $= (d y / d u) (d u / d x)$ $= 2 u \cdot 3$ . Each unit of change in $x$ makes $3$ units of change in $u$ , and each unit of change in $u$ makes $2 u$ units of change in $y$ .

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

Put the inside back

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

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

Simplify

Multiply the numbers first, $2 \cdot 3 = 6$ , so $d y / d x$ $= 6 (3 x + 1)$ $= 18 x + 6$ .

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

Check by expanding

${(3 x + 1)}^{2} = 9 x^{2} + 6 x + 1$ , and the power rule gives $18 x + 6$ . The two routes agree. For higher powers, such as ${(3 x + 1)}^{50}$ , 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

$d y / d x$ $= 6 (3 x + 1)$ $= 18 x + 6$ .

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

Your turn

1. One like the example

Differentiate $f (x) = {(2 x - 3)}^{3}$ .

Show the answer and explanation

$f^{'} (x) = 6 {(2 x - 3)}^{2}$ .

The outer function cubes, so its derivative is $3 u^{2}$ at $u = 2 x - 3$ . Multiply by the inner derivative $2$ to get $f^{'} (x)$ $= 3 {(2 x - 3)}^{2} \cdot 2$ $= 6 {(2 x - 3)}^{2}$ .

\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}

2. A variation

Differentiate $f (x) = {(5 - x^{2})}^{4}$ . This time the inside is not a straight line.

Show the answer and explanation

$f^{'} (x) = - 8 x {(5 - x^{2})}^{3}$ .

The outside is the 4th power, so its derivative is $4 u^{3}$ at $u = 5 - x^{2}$ . The inside, $5 - x^{2}$ , has derivative $- 2 x$ . Multiply to get $4 {(5 - x^{2})}^{3} \cdot (- 2 x)$ $= - 8 x {(5 - x^{2})}^{3}$ .

\begin{array}{l} f (x) = {(5 - x^{2})}^{4} \\ f^{'} (x) = 4 {(5 - x^{2})}^{3} \cdot (- 2 x) \\ f^{'} (x) = - 8 x {(5 - x^{2})}^{3} \end{array}

3. Somewhere new

A cube’s side is $2 t + 1 cm$ after $t$ seconds, so its volume is $V (t) = {(2 t + 1)}^{3}$ ${cm}^{3}$ . How fast is the volume growing at $t = 1$ ?

Show the answer and explanation

$V^{'} (1) = 54 {cm}^{3}$ per second.

$V$ is the cube of the inside $2 t + 1$ . The outside gives $3 {(2 t + 1)}^{2}$ and the inside gives $2$ , so $V^{'} (t) = 6 {(2 t + 1)}^{2}$ . At $t = 1$ the side is $3 cm$ , and $V^{'} (1) = 6 \cdot 9 = 54 {cm}^{3}$ per second.

\begin{array}{l} V (t) = {(2 t + 1)}^{3} \\ V^{'} (t) = 3 {(2 t + 1)}^{2} \cdot 2 \\ V^{'} (t) = 6 {(2 t + 1)}^{2} \\ V^{'} (1) = 54 \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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Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 4 key terms Open Chain Rule Composition Check the derivative in Math Open worked example on a board Chain rule in Math Reference

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Name the inside and the outside

Multiply the rates

Put the inside back

Simplify

Check by expanding

Result

Your turn

1. One like the example

f′(x)=6⁢(2⁢x−3)2f′(x)=6(2x−3)2.

2. A variation

f′(x)=−8⁢x⁢(5−x2)3f′(x)=−8x(5−x2)3.

3. Somewhere new

V′(1)=54 cm3V′(1)=54 cm3 per second.

Keep exploring

Study it, then try it

$f^{'} (x) = 6 {(2 x - 3)}^{2}$ .

$f^{'} (x) = - 8 x {(5 - x^{2})}^{3}$ .

$V^{'} (1) = 54 {cm}^{3}$ per second.