9B - The chain rule

How can you differentiate a bracket raised to a power without expanding it? You will separate an inner function from an outer function, then use the chain rule to differentiate powers, trigonometric functions and exponentials of simple expressions.

A function of a function

To find the gradient of $y=(3x-2)^4$ , must you multiply out four brackets first?

Think about how you calculate its value. First calculate $3x-2$ ; then raise that answer to the fourth power. At $x=1$ , these steps give $1$ , then $1$ . At $x=2$ , they give $4$ , then $256$ . The fourth power acts on the whole result of the first calculation.

Composite function

A composite function applies one function to the output of another. If $u=g(x)$ and $y=f(u)$ , then $y=f(g(x))$ . We call $g$ the inner function and $f$ the outer function.

Read the operations from the input outwards. The outer function is the last operation performed.

Expression	Inner function $u$	Outer function of $u$
$(3x-2)^4$	$3x-2$	$u^4$
$\sqrt{5-x}$	$5-x$	$\sqrt{u}$
$\sin(x^2)$	$x^2$	$\sin u$
$e^{1-x^2}$	$1-x^2$	$e^u$

The distinction affects differentiation. In $\sin(x^2)$ , the sine acts on $x^2$ . In $(\sin x)^2$ , squaring acts on $\sin x$ . They have different inner and outer functions. Likewise, $(3x-2)^4$ is different from $3x^4-2$ , where the fourth power applies only to $x$ .

Key idea

Identify the last operation first. Temporarily replacing the inner expression by $u$ makes the outer function easier to recognise.

Use that distinction in the next check.

Applying the chain rule

The outer function changes because its input changes, so both rates contribute to the derivative.

For $y=(3x-2)^4$ , put $u=3x-2$ . At $x=1$ , we have $u=1$ . Near this point, $u$ changes three times as fast as $x$ , while $u^4$ changes four times as fast as $u$ . The resulting rate is $4\times3=12$ units of $y$ per unit of $x$ .

Chain rule

If $u=g(x)$ is differentiable at the point considered and $y=f(u)$ is differentiable at the corresponding value of $u$ , then

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} =f'(g(x))g'(x).

Differentiate the outer function, keeping its inner expression, then multiply by the derivative of the inner function.

Not on the formulae sheet: learn it.

For a power, this gives $\frac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x)$ , wherever the functions involved have the required real derivatives. The power falls by one; the inner expression itself stays intact.

Worked example 1: a power of a bracket

Question. Differentiate $y=(3x-2)^4$ .

Plan. Use $u=3x-2$ , differentiate $u^4$ , then include the rate of change of $u$ .

\begin{aligned} u&=3x-2 &\frac{du}{dx}&=3\\ y&=u^4 &\frac{dy}{du}&=4u^3\\ \frac{dy}{dx}&=4u^3\times3 &&\text{multiply the two rates}\\ &=12(3x-2)^3 &&\text{replace }u \end{aligned}

Answer. $\frac{dy}{dx}=12(3x-2)^3$ , for all real $x$ .

Check. Expanding the original gives $81x^4-216x^3+216x^2-96x+16$ . Its derivative is $324x^3-648x^2+432x-96$ , which is the expansion of $12(3x-2)^3$ .

Keep the same inner expression but change the outer power to a square root. The chain-rule steps stay the same; the real domain now matters.

Worked example 2: a square root of a bracket

Question. Differentiate $y=\sqrt{3x-2}$ and state where the derivative formula applies.

Plan. Write the root as a power of $u=3x-2$ .

\begin{aligned} y&=u^{1/2} &\frac{dy}{du}&=\tfrac12u^{-1/2}\\ \frac{dy}{dx}&=\tfrac12(3x-2)^{-1/2}\times3 &&\text{include the inner derivative}\\ &=\frac{3}{2\sqrt{3x-2}} &&\text{write with a positive power} \end{aligned}

The function is real for $x\ge\frac23$ . Its displayed derivative is finite only for $x>\frac23$ ; it does not give a finite derivative at the endpoint.

Answer. $\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}$ , for $x>\frac23$ .

Check. At $x=1$ , the formula gives $1.5$ . Using the original function, the gradient between $x=0.9999$ and $x=1.0001$ is approximately $1.50000002$ , consistent with this value.

The same rule also handles negative powers, provided the original denominator is nonzero.

Watch out

$\frac{d}{dx}(3x-2)^4=4(3x-2)^3$ misses the factor $3$ . The outer derivative is only one of the two rates.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Series

Geometry

Trigonometry

Calculus

9B - The chain rule

A function of a function

Applying the chain rule

Continue this lesson.

Learn faster with Nova