The chain rule

AS · P1 · 11 min

The power rule differentiates $x^5$ , but what about $(3x - 2)^5$ or $\sqrt{2x^3 + 5}$ ? These are composite functions: one expression sits inside another. You could expand $(3x - 2)^5$ , but you cannot expand a square root, and the chain rule handles both in one line. In Paper 1 it appears in almost every calculus question, inside tangents, stationary points and rates of change, so it needs to be automatic.

Functions inside functions

In $y = (3x - 2)^5$ there are two steps: first work out the inner function $u = 3x - 2$ , then apply the outer function, raising to the power $5$ . This is exactly the composite function idea from the functions unit: $y = f(g(x))$ with $g(x) = 3x - 2$ and $f(u) = u^5$ .

Seeing the pattern

Take $y = (2x + 1)^2$ , which you can expand:

y = 4x^2 + 4x + 1 \quad\Rightarrow\quad \frac{dy}{dx} = 8x + 4 = 4(2x + 1)

Now compare with treating the bracket as if it were a single letter. The power rule on $u^2$ gives $2u = 2(2x + 1)$ . That is only half of the right answer. The missing factor is $2$ , which is the derivative of the inside, $2x + 1$ :

\frac{dy}{dx} = \underbrace{2(2x + 1)}_{\text{outer, inside left alone}} \times \underbrace{2}_{\text{derivative of inside}} = 4(2x + 1)

The extra factor is there because the inside changes at its own rate. If $x$ increases by a small amount, $u = 2x + 1$ increases twice as fast, and $y$ responds to the change in $u$ . Rates multiply.

The rule

Key result

If $y$ is a function of $u$ , and $u$ is a function of $x$ , then

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

For a power of a bracket this becomes

\frac{d}{dx}\Big[\big(f(x)\big)^n\Big] = n\big(f(x)\big)^{n - 1} \times f'(x)

and in the most common special case, a linear inside,

\frac{d}{dx}\Big[(ax + b)^n\Big] = an(ax + b)^{n - 1}

In words: differentiate the outside, leaving the inside alone, then multiply by the derivative of the inside. The rule works for any rational power $n$ , so it covers roots and reciprocals of brackets once you rewrite them as powers.

Using the chain rule

Rewrite the expression as $(\text{inside})^n$ . Roots become fractional powers and reciprocals become negative powers: $\sqrt{2x^3 + 5} = (2x^3 + 5)^{\frac{1}{2}}$ and $\dfrac{8}{(2x + 1)^2} = 8(2x + 1)^{-2}$ .
Bring the power down and reduce it by one, copying the inside unchanged.
Multiply by the derivative of the inside.
Tidy up: collect numerical factors, and rewrite negative and fractional powers if a value is to be substituted.

On paper you can set out the substitution explicitly, which is good practice until it is automatic:

u = 2x^3 + 5,\quad y = u^{\frac{1}{2}} \qquad \frac{du}{dx} = 6x^2,\quad \frac{dy}{du} = \tfrac{1}{2}u^{-\frac{1}{2}}

\frac{dy}{dx} = \tfrac{1}{2}u^{-\frac{1}{2}} \times 6x^2 = \frac{3x^2}{\sqrt{2x^3 + 5}}

That is the exact example the syllabus gives.

What the graph shows

y = (3x - 2)^2 y = 6x - 5 (1, 1)

$y = (3x - 2)^2$ is the parabola $y = x^2$ squeezed horizontally by a factor $3$ and shifted right. Squeezing a curve horizontally by a factor of $3$ makes it three times as steep, and that is the factor $3$ the chain rule brings in. At $(1, 1)$ the derivative $6(3x - 2)$ gives gradient $6$ , and the tangent is $y = 6x - 5$ . Without the factor $3$ you would get gradient $2$ , a line that visibly does not touch the curve.

Second derivatives

To find $\dfrac{d^2y}{dx^2}$ apply the chain rule again to $\dfrac{dy}{dx}$ . For a linear inside, each differentiation brings out another factor $a$ :

y = (2x - 3)^3 \quad\Rightarrow\quad \frac{dy}{dx} = 6(2x - 3)^2 \quad\Rightarrow\quad \frac{d^2y}{dx^2} = 12(2x - 3) \times 2 = 24(2x - 3)

Worked examples

A linear inside

Find $\dfrac{dy}{dx}$ when $y = (3x - 2)^5$ , and find the gradient of the curve at the point where $x = 1$ .

Solution

Outer function $u^5$ , inner $u = 3x - 2$ with derivative $3$ :

\frac{dy}{dx} = 5(3x - 2)^4 \times 3 = 15(3x - 2)^4

At $x = 1$ : $15(1)^4 = 15$ .

A square root of a cubic

Find $\dfrac{dy}{dx}$ when $y = \sqrt{2x^3 + 5}$ , and find the exact gradient of the curve at the point where $x = -1$ .

Solution

$y = (2x^3 + 5)^{\frac{1}{2}}$ , so

\frac{dy}{dx} = \tfrac{1}{2}(2x^3 + 5)^{-\frac{1}{2}} \times 6x^2 = \frac{3x^2}{\sqrt{2x^3 + 5}}

At $x = -1$ : $2x^3 + 5 = -2 + 5 = 3$ and $3x^2 = 3$ , so the gradient is

\frac{3}{\sqrt{3}} = \sqrt{3}

A reciprocal, with the second derivative

A curve has equation $y = \dfrac{8}{(2x + 1)^2}$ . Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ , and evaluate both at the point where $x = \tfrac{1}{2}$ .

Solution

$y = 8(2x + 1)^{-2}$ .

\frac{dy}{dx} = 8 \times (-2)(2x + 1)^{-3} \times 2 = -32(2x + 1)^{-3}

\frac{d^2y}{dx^2} = -32 \times (-3)(2x + 1)^{-4} \times 2 = 192(2x + 1)^{-4}

At $x = \tfrac{1}{2}$ , $2x + 1 = 2$ :

\frac{dy}{dx} = \frac{-32}{8} = -4, \qquad \frac{d^2y}{dx^2} = \frac{192}{16} = 12

Finding where the gradient is zero

Find the $x$ -coordinates of the points on $y = (x^2 - 4x + 1)^3$ where the gradient is zero, giving exact answers.

Solution

The inside is $x^2 - 4x + 1$ with derivative $2x - 4$ :

\frac{dy}{dx} = 3(x^2 - 4x + 1)^2(2x - 4) = 6(x - 2)(x^2 - 4x + 1)^2

This is zero when $x - 2 = 0$ or $x^2 - 4x + 1 = 0$ . The quadratic gives

x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}

So the gradient is zero at $x = 2$ , $x = 2 + \sqrt{3}$ and $x = 2 - \sqrt{3}$ .

Do not expand the bracket: the factorised derivative is what lets you solve $\dfrac{dy}{dx} = 0$ .

Unknown constants inside a root

The curve $y = \sqrt{ax + b}$ , where $a$ and $b$ are positive constants, passes through $(2, 3)$ and has gradient $\tfrac{1}{3}$ there. Find $a$ and $b$ .

Solution

On the curve: $3 = \sqrt{2a + b}$ , so

2a + b = 9 \qquad (1)

Differentiate $y = (ax + b)^{\frac{1}{2}}$ :

\frac{dy}{dx} = \tfrac{1}{2}(ax + b)^{-\frac{1}{2}} \times a = \frac{a}{2\sqrt{ax + b}}

At $(2, 3)$ the root $\sqrt{2a + b}$ equals $3$ , so

\frac{a}{2 \times 3} = \frac{1}{3} \quad\Rightarrow\quad a = 2

Then from (1), $b = 9 - 4 = 5$ . The curve is $y = \sqrt{2x + 5}$ .

Two chain rule terms

The function $f$ is defined by $f(x) = \sqrt{2x - 1} + \dfrac{4}{2x - 1}$ for $x > \tfrac{1}{2}$ .

(a) Find $f'(x)$ .

(b) Find the coordinates of the point on $y = f(x)$ at which the gradient is zero.

Solution

(a) Write $f(x) = (2x - 1)^{\frac{1}{2}} + 4(2x - 1)^{-1}$ . Both terms have inside $2x - 1$ with derivative $2$ :

f'(x) = \tfrac{1}{2}(2x - 1)^{-\frac{1}{2}} \times 2 + 4(-1)(2x - 1)^{-2} \times 2 = (2x - 1)^{-\frac{1}{2}} - 8(2x - 1)^{-2}

(b) Set $f'(x) = 0$ :

(2x - 1)^{-\frac{1}{2}} = 8(2x - 1)^{-2}

Multiply both sides by $(2x - 1)^2$ (positive, since $x > \tfrac{1}{2}$ ):

(2x - 1)^{\frac{3}{2}} = 8 \quad\Rightarrow\quad 2x - 1 = 8^{\frac{2}{3}} = 4 \quad\Rightarrow\quad x = \tfrac{5}{2}

Then $f\!\left(\tfrac{5}{2}\right) = \sqrt{4} + \dfrac{4}{4} = 2 + 1 = 3$ . The point is $\left(\tfrac{5}{2}, 3\right)$ .

When a whole expression is built from the same bracket, treat the bracket as one quantity. Here, writing $u = 2x - 1$ turns the equation into $u^{-\frac{1}{2}} = 8u^{-2}$ , which is much easier to see.

Watch out

Forgetting to multiply by the derivative of the inside. $\dfrac{d}{dx}(3x - 2)^5 = 5(3x - 2)^4$ is wrong; the factor $3$ is missing. This is the most common chain rule error, and it loses the accuracy mark every time.

Changing the inside. The bracket is copied unchanged: $\dfrac{d}{dx}(x^2 + 1)^4 = 4(x^2 + 1)^3 \times 2x$ , not $4(2x)^3$ .

Losing a negative sign from the inside. $\dfrac{d}{dx}(5 - 2x)^6 = 6(5 - 2x)^5 \times (-2) = -12(5 - 2x)^5$ .

Mishandling the reciprocal. $\dfrac{4}{(3x + 2)^2}$ is $4(3x + 2)^{-2}$ , not $(4(3x + 2))^{-2}$ and not $\tfrac{1}{4}(3x + 2)^{-2}$ .

Expanding when you should not. Expanding $(x^2 - 4x + 1)^3$ is slow and destroys the factorised form you need to solve $\dfrac{dy}{dx} = 0$ .

Tip

Because $\dfrac{dy}{dx}$ behaves like a fraction in the chain rule, you also have $\dfrac{dy}{dx} = 1 \Big/ \dfrac{dx}{dy}$ . This is useful in connected rates of change, where you often need $\dfrac{dr}{dV}$ from a formula that gives $V$ in terms of $r$ .

Exam tip

Marks for the chain rule are typically one for the correct power and bracket, and one for the factor from the inside. Write the factor explicitly, as in $5(3x - 2)^4 \times 3$ , before simplifying, so that a numerical slip still earns the method mark.
"Find $\dfrac{dy}{dx}$ " answers do not need to be fully simplified, but they must be correct. A quick check: the power of the bracket should go down by exactly one.
Substituting values: work out the inside first ( $2x + 1 = 2$ at $x = \tfrac{1}{2}$ ), then apply powers. It is faster and avoids calculator errors with fractional powers.
Exact answers: surds such as $\sqrt{3}$ or $\dfrac{3}{\sqrt{7}}$ are exact; do not convert them to decimals unless asked.

Summary

A composite function has an inner function inside an outer one; the chain rule is $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ .
For powers of a bracket: $\dfrac{d}{dx}\big[(f(x))^n\big] = n\big(f(x)\big)^{n-1}f'(x)$ .
Linear inside: $\dfrac{d}{dx}(ax + b)^n = an(ax + b)^{n-1}$ .
Rewrite roots and reciprocals of brackets as powers before differentiating.
Copy the inside unchanged and multiply by its derivative, including any minus sign.
For a second derivative, apply the chain rule again.
Keep derivatives factorised when you need to solve $\dfrac{dy}{dx} = 0$ .

Practice questions

Question

Differentiate $(4x + 1)^3$ .
Differentiate $(5 - 2x)^6$ .
Find the gradient of the curve $y = \sqrt{6x - 5}$ at the point where $x = 1$ .
Find the gradient of $y = \dfrac{4}{(3x + 2)^2}$ at the point where it crosses the $y$ -axis.
Find $\dfrac{dy}{dx}$ when $y = (x^2 + 1)^4$ .
Find the exact gradient of $y = \dfrac{1}{\sqrt{x^3 + 8}}$ at the point where $x = 2$ .
Given $f(x) = (2x - 3)^3$ , find $f'(x)$ and $f''(x)$ , and evaluate $f''(2)$ .
Find the coordinates of the points on $y = (x^2 - 9)^2$ where the gradient is zero.
The function $f$ is defined by $f(x) = 2\sqrt{3x + 1} - x$ for $x > -\tfrac{1}{3}$ . Find the coordinates of the point on $y = f(x)$ where the gradient is zero, and the value of $f''(x)$ at that point.
The gradient of the curve $y = \sqrt{x^2 + k}$ at the point where $x = 2$ is $\tfrac{2}{3}$ . Find the value of the constant $k$ , and hence find the equation of the tangent at that point in the form $y = mx + c$ .

Answers

$3(4x + 1)^2 \times 4 = 12(4x + 1)^2$ .
$6(5 - 2x)^5 \times (-2) = -12(5 - 2x)^5$ .
$y = (6x - 5)^{\frac{1}{2}}$ , so $\dfrac{dy}{dx} = \tfrac{1}{2}(6x - 5)^{-\frac{1}{2}} \times 6 = \dfrac{3}{\sqrt{6x - 5}}$ . At $x = 1$ : $\dfrac{3}{\sqrt{1}} = 3$ .
$y = 4(3x + 2)^{-2}$ , $\dfrac{dy}{dx} = -8(3x + 2)^{-3} \times 3 = -24(3x + 2)^{-3}$ . It crosses the $y$ -axis at $x = 0$ : gradient $\dfrac{-24}{8} = -3$ .
$4(x^2 + 1)^3 \times 2x = 8x(x^2 + 1)^3$ .
$y = (x^3 + 8)^{-\frac{1}{2}}$ , $\dfrac{dy}{dx} = -\tfrac{1}{2}(x^3 + 8)^{-\frac{3}{2}} \times 3x^2 = -\dfrac{3x^2}{2(x^3 + 8)^{3/2}}$ . At $x = 2$ : $x^3 + 8 = 16$ and $16^{\frac{3}{2}} = 64$ , so the gradient is $-\dfrac{12}{128} = -\dfrac{3}{32}$ .
$f'(x) = 3(2x - 3)^2 \times 2 = 6(2x - 3)^2$ . $f''(x) = 12(2x - 3) \times 2 = 24(2x - 3)$ . $f''(2) = 24(1) = 24$ .
$\dfrac{dy}{dx} = 2(x^2 - 9) \times 2x = 4x(x - 3)(x + 3)$ . Zero at $x = 0, 3, -3$ . The points are $(0, 81)$ , $(3, 0)$ and $(-3, 0)$ .
$f(x) = 2(3x + 1)^{\frac{1}{2}} - x$ , so $f'(x) = (3x + 1)^{-\frac{1}{2}} \times 3 - 1 = \dfrac{3}{\sqrt{3x + 1}} - 1$ . Setting this to zero: $\sqrt{3x + 1} = 3$ , so $3x + 1 = 9$ and $x = \tfrac{8}{3}$ . Then $f\!\left(\tfrac{8}{3}\right) = 2(3) - \tfrac{8}{3} = \tfrac{10}{3}$ . The point is $\left(\tfrac{8}{3}, \tfrac{10}{3}\right)$ . Next, $f''(x) = 3 \times \left(-\tfrac{1}{2}\right)(3x + 1)^{-\frac{3}{2}} \times 3 = -\tfrac{9}{2}(3x + 1)^{-\frac{3}{2}}$ . At $x = \tfrac{8}{3}$ , $(3x + 1)^{\frac{3}{2}} = 9^{\frac{3}{2}} = 27$ , so $f'' = -\dfrac{9}{54} = -\dfrac{1}{6}$ .
$y = (x^2 + k)^{\frac{1}{2}}$ , $\dfrac{dy}{dx} = \tfrac{1}{2}(x^2 + k)^{-\frac{1}{2}} \times 2x = \dfrac{x}{\sqrt{x^2 + k}}$ . At $x = 2$ : $\dfrac{2}{\sqrt{4 + k}} = \dfrac{2}{3}$ , so $\sqrt{4 + k} = 3$ and $k = 5$ . The point is $(2, \sqrt{9}) = (2, 3)$ , so the tangent is $y - 3 = \tfrac{2}{3}(x - 2)$ , i.e. $y = \tfrac{2}{3}x + \tfrac{5}{3}$ .

Functions inside functions

Seeing the pattern

The rule

What the graph shows

Second derivatives

Worked examples

Practice questions

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