1.7 - Differentiating a function of a function | Cambridge A-Level Mathematics

The expression inside a power may itself change with $x$ . You will use the chain rule to follow both stages of that change, including roots and reciprocal powers with nonlinear expressions inside. Expanding simple cases will help you check that no factor has been missed.

The inner and outer functions

Why is the derivative of $(3x+1)^4$ not just $4(3x+1)^3$ ? The fourth-power operation acts on $3x+1$ , and that inner expression changes three times as fast as $x$ .

Break the calculation into two stages: $u=3x+1$ , then $y=u^4$ . Near a particular input, a small change in $x$ produces a change in $u$ , which in turn produces a change in $y$ . The two local rates multiply.

Inner and outer functions

In $y=[g(x)]^n$ , the inner function is $u=g(x)$ , and the outer function is $y=u^n$ . A composite function applies the inner function first, then the outer function.

Keep the expression inside the power intact while differentiating the outer operation.

The chain rule

Where the required derivatives exist,

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

In particular, for constant rational $n$ ,

\frac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x).

Apply this where the original function and the derivative are defined.

Not stated separately in the formula booklet: learn it.

The notation resembles cancelling $du$ , which is a helpful memory aid. The rule expresses the multiplication of local rates; the symbols are derivative notation, not ordinary fractions that can always be rearranged without checking conditions.

Worked example 1: an inner rate of one

Question. Differentiate $y=(x+1)^4$ .

Plan. Differentiate the fourth power, then multiply by the derivative of $x+1$ .

\begin{aligned} u&=x+1, &\frac{du}{dx}&=1\\ \frac{dy}{du}&=4u^3, &\frac{dy}{dx}&=4(x+1)^3\times1. \end{aligned}

Answer. $dy/dx=4(x+1)^3$ .

Check. Expanding gives $x^4+4x^3+6x^2+4x+1$ , whose derivative is $4x^3+12x^2+12x+4=4(x+1)^3$ .

Change only the coefficient of $x$ inside the bracket. The inner rate is now $3$ .

Worked example 2: an inner rate of three

Question. Differentiate $y=(3x+1)^4$ .

Plan. Use the same outer derivative and multiply by the new inner derivative.

\begin{aligned} u&=3x+1, &\frac{du}{dx}&=3\\ \frac{dy}{dx}&=4u^3\times3\\ &=12(3x+1)^3. \end{aligned}

Answer. $dy/dx=12(3x+1)^3$ .

Check. At $x=0$ , the derivative is $12$ . Expanding the original expression gives a linear term $12x$ ; every higher-power term has derivative zero at $0$ , so the result agrees.

The opening question is resolved: $4(3x+1)^3$ measures change per unit change in the inner input; multiplying by $3$ converts it to change per unit $x$ .

Watch out

$d[(3x+1)^4]/dx=12(3x+1)^3$ , not $4(3x+1)^3$ . Write the inner derivative as a separate factor before simplifying.

Try another linear inner function.

Nonlinear inner functions

The inner function does not have to be linear. For $y=\sqrt{x^3+4}$ , the inside is $u=x^3+4$ , whose rate of change is $3x^2$ . Rewrite the outer root as a power before using the same two-stage method.

For a square root, the real expression needs $x^3+4\ge0$ . The derivative formula obtained below has a square root in the denominator, so it applies where $x^3+4>0$ . The endpoint where the root is zero needs separate attention; do not substitute it into a denominator.

Worked example 3: a root with a cubic inside

Question. Differentiate $y=\sqrt{x^3+4}$ , where $x^3+4>0$ , and find the derivative at $x=2$ .

Plan. Use outer exponent $1/2$ and inner derivative $3x^2$ .

\begin{aligned} y&=(x^3+4)^{1/2}\\ \frac{dy}{dx}&=\frac12(x^3+4)^{-1/2}\times3x^2\\ &=\frac{3x^2}{2\sqrt{x^3+4}}\\ y'(2)&=\frac{12}{2\sqrt{12}}=\sqrt3. \end{aligned}

Answer. $dy/dx=3x^2/(2\sqrt{x^3+4})$ , with $y'(2)=\sqrt3$ .

Check. At $x=2$ , the inside changes at rate $12$ , while the square-root operation changes at rate $1/(2\sqrt{12})$ . Their product is $\sqrt3\approx1.73205$ . A chord to $2.001$ gives approximately $1.73248$ .

Keep the same inner function and move the root to the denominator. Only the outer exponent changes, from $1/2$ to $-1/2$ .

Worked example 4: a reciprocal root

Question. Differentiate $y=1/\sqrt{x^3+4}$ , where $x^3+4>0$ , and find the derivative at $x=2$ .

Plan. Apply the power rule with exponent $-1/2$ , then multiply by $3x^2$ .

\begin{aligned} y&=(x^3+4)^{-1/2}\\ \frac{dy}{dx}&=-\frac12(x^3+4)^{-3/2}\times3x^2\\ &=-\frac{3x^2}{2(x^3+4)^{3/2}}\\ y'(2)&=-\frac{12}{2(12)^{3/2}}=-\frac1{4\sqrt3}. \end{aligned}

Answer. $dy/dx=-3x^2/[2(x^3+4)^{3/2}]$ , with $y'(2)=-1/(4\sqrt3)$ .

Check. The positive inner expression increases near $2$ ; its reciprocal root decreases. The negative sign fits. The chord to $2.001$ has gradient approximately $-0.144301$ , close to $-1/(4\sqrt3)\approx-0.144338$ .

For an integer negative outer power, only a nonzero inner value is required. For example, $(x^2-9)^{-2}$ is defined when $x\ne\pm3$ , and its derivative is $-4x(x^2-9)^{-3}$ . Always read the domain from the complete expression.

Problem-solving

Write three things before simplifying: the outer exponent, the unchanged inner expression, and the inner derivative. Then combine coefficients and rewrite negative powers if useful.

Apply the root method to a new cubic inside.

Checking the chain factor

An expanded special case gives an independent check on a chain-rule result. This is particularly useful when the inside contains a coefficient or a power of $x$ . Choose expansion when it is short; keep the composite form when expansion would create many terms or cannot remove the root.

Worked example 5: checking a linear inner function

Question. Find the derivative of $y=(2x-1)^2$ in two ways.

Plan. Differentiate by the chain rule, then expand the original square and differentiate term by term.

\begin{aligned} \text{Chain rule:}\quad y'&=2(2x-1)\times2=8x-4\\ \text{Expansion:}\quad y&=4x^2-4x+1\\ y'&=8x-4. \end{aligned}

Answer. Both methods give $8x-4$ .

Check. At $x=1$ , both give gradient $4$ . Omitting the inner factor would give $2$ , contradicting the expanded result.

Change only the inner function to a quadratic. The extra chain factor now varies with $x$ .

Worked example 6: checking a nonlinear inner function

Question. Find the derivative of $y=(x^2-1)^2$ in two ways.

Plan. Keep the inner derivative $2x$ in the chain-rule method, then compare with expansion.

\begin{aligned} \text{Chain rule:}\quad y'&=2(x^2-1)(2x)\\ &=4x(x^2-1)\\ \text{Expansion:}\quad y&=x^4-2x^2+1\\ y'&=4x^3-4x. \end{aligned}

Answer. $y'=4x(x^2-1)=4x^3-4x$ .

Check. At $x=2$ , the inner rate is $4$ , and the outer rate is $2(4-1)=6$ ; their product is $24$ . The expanded derivative gives $32-8=24$ .

You can also check the structure without expanding: differentiating a function of $x^2$ usually introduces a factor of $x$ . This is a useful warning signal, although agreement at a few inputs is not a proof of an identity.

Watch out

For $y=(x^2-1)^2$ , the line $y'=2(x^2-1)$ has differentiated only the outer function. Multiply by $2x$ , not by $x^2-1$ again.

Find and repair an omitted factor in a new expression.

Quick Recap

Identify the inner function and the outer operation.
Differentiate the outer operation, leaving its input intact, then multiply by the inner derivative.
Rewrite roots and reciprocal powers before using the rule.
Check domains and denominators before evaluating a derivative.
Use a short expansion as an independent check when possible.

The missing factor in the opening question is $3$ : $d[(3x+1)^4]/dx=12(3x+1)^3$ .

Last Accessed

Pure Mathematics 1

Advanced pure mathematics

Pure Mathematics 3 extensions

Pure Mathematics 2 only

Mechanics

Probability and Statistics 1

Probability and Statistics 2

1.7 - Differentiating a function of a function

The inner and outer functions

Nonlinear inner functions

Checking the chain factor

Quick Recap

Learn faster with Nova