Year 11 Methods (Unit 1 & 2) Further Differentiation

The Chain Rule

Q: What is the chain rule?

If y is a function of u and u is a function of x, then dy dx equals dy du times du dx. You differentiate the outer function and multiply by the derivative of the inner function.

Q: How do you differentiate a bracket raised to a power?

Bring the power down and reduce it by one, then multiply by the derivative of the inside. The derivative of g of x to the n is n times g of x to the n minus one, times g prime of x.

Q: Why is my chain rule answer wrong by a constant factor?

You have almost certainly left out the inner derivative g prime of x. For three x plus two all to the fourth, the missing factor is the times three.

Q: How do you use the chain rule on a square root?

Rewrite the square root of g of x as g of x to the power one half, then differentiate to get one half times g of x to the power minus one half, times g prime of x.

Q: What are the inner and outer functions?

The inner function is the expression inside the bracket, u equals g of x. The outer function is what is done to it, here raising to a power, y equals u to the n.

20 practice questions 1 video lesson Theory + worked examples

Master the chain rule for Queensland Year 11 Mathematical Methods (QCAA). It differentiates a composite function — a function inside another function — by differentiating the outer part and then multiplying by the derivative of the inside.

You will learn to spot the inner and outer functions, differentiate powers of polynomials, handle roots and reciprocals by rewriting them first, and give the derivative in simplest and factorised form.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the chain rule differentiates a composite function: if \(y=f(u)\) and \(u=g(x)\) then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). This page applies it to powers of polynomials such as \((ax+b)^{n}\) and \((\text{polynomial})^{n}\) — including reciprocal and root forms once they are rewritten as powers.

A composite function is a function of a function: an inner function \(u=g(x)\) sitting inside an outer function \(y=f(u)\). For example \(y=(3x+2)^{4}\) has inner \(u=3x+2\) and outer \(y=u^{4}\).

The chain rule says the derivative is the outer derivative times the inner derivative: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). You differentiate the outer power (leaving the inner bracket untouched), then multiply by the derivative of the inside.

Reciprocal and root composites are handled by rewriting first: \(\dfrac{1}{(\ldots)^{n}}=(\ldots)^{-n}\) and \(\sqrt{\ldots}=(\ldots)^{1/2}\), then apply the chain rule as usual.

Outer × inner. Differentiate the bracket to the power (\(n\,u^{\,n-1}\)) and then multiply by the derivative of what is inside the bracket. Missing the inner factor is the number-one slip.

\(y=(x^{2}-1)^{2}\): an inner polynomial inside an outer square.

\(y=\sqrt{x^{2}+9}=(x^{2}+9)^{1/2}\): a root composite, rewritten as a power.

The chain rule, with \(y=f(u)\) and \(u=g(x)\):

\[\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\]

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

For a bracket raised to a power, this becomes the useful shortcut:

\[\dfrac{d}{dx}\big(g(x)\big)^{n}=n\,\big(g(x)\big)^{n-1}\cdot g^{\prime}(x)\]

\frac{d}{d x} {(g (x))}^{n} = n {(g (x))}^{n - 1} \cdot g^{'} (x)

Do not forget \(g^{\prime}(x)\). The inner derivative is the factor most often left out; for \((3x+2)^{4}\) it supplies the extra \(\times 3\).

How to use the chain rule

Name the inside: let \(u=g(x)\) be the inner function, so \(y=u^{n}\). Rewrite any root or reciprocal as a power first.
Differentiate each part: \(\dfrac{dy}{du}=n\,u^{n-1}\) and \(\dfrac{du}{dx}=g'(x)\).
Multiply and back-substitute: \(\dfrac{dy}{dx}=n\,u^{n-1}\cdot g'(x)\), then replace \(u\) with \(g(x)\) and tidy.

Example 1 — A linear inside a power

Differentiate \(y=(3x+2)^{4}\).

Solution

Name the inside and outside:

\(u\)	\(=\)	\(3x+2\)
\(y\)	\(=\)	\(u^{4}\)

Differentiate each part:

\(\dfrac{dy}{du}\)	\(=\)	\(4u^{3}\)
\(\dfrac{du}{dx}\)	\(=\)	\(3\)

Multiply (outer \(\times\) inner) and back-substitute:

\(\dfrac{dy}{dx}\)	\(=\)	\(4u^{3}\times 3\)
	\(=\)	\(12u^{3}\)
	\(=\)	\(12(3x+2)^{3}\)

\(\dfrac{dy}{dx}=12(3x+2)^{3}\).

12 {(3 x + 2)}^{3}

Example 2 — A polynomial inside a power

Differentiate \(y=(x^{2}-5)^{3}\).

Solution

Name the inside and outside:

\(u\)	\(=\)	\(x^{2}-5\)
\(y\)	\(=\)	\(u^{3}\)

Differentiate each part:

\(\dfrac{dy}{du}\)	\(=\)	\(3u^{2}\)
\(\dfrac{du}{dx}\)	\(=\)	\(2x\)

Multiply and back-substitute:

\(\dfrac{dy}{dx}\)	\(=\)	\(3u^{2}\times 2x\)
	\(=\)	\(6x\,u^{2}\)
	\(=\)	\(6x(x^{2}-5)^{2}\)

\(\dfrac{dy}{dx}=6x(x^{2}-5)^{2}\).

6 x {(x^{2} - 5)}^{2}

Example 3 — A reciprocal composite

Differentiate \(y=\dfrac{1}{(2x-1)^{2}}\).

Solution

Rewrite with a negative power, then name the inside:

\(y\)	\(=\)	\((2x-1)^{-2}\)
\(u\)	\(=\)	\(2x-1\)

Differentiate each part:

\(\dfrac{dy}{du}\)	\(=\)	\(-2u^{-3}\)
\(\dfrac{du}{dx}\)	\(=\)	\(2\)

Multiply, back-substitute and tidy:

\(\dfrac{dy}{dx}\)	\(=\)	\(-2u^{-3}\times 2\)
	\(=\)	\(-4(2x-1)^{-3}\)
	\(=\)	\(-\dfrac{4}{(2x-1)^{3}}\)

\(\dfrac{dy}{dx}=-\dfrac{4}{(2x-1)^{3}}\).

- \frac{4}{{(2 x - 1)}^{3}}

Example 4 — A root composite, evaluated

For \(y=\sqrt{x^{2}+9}\), find \(\dfrac{dy}{dx}\) and its value at \(x=4\).

Solution

Rewrite the root as a power and name the inside:

\(y\)	\(=\)	\((x^{2}+9)^{1/2}\)
\(u\)	\(=\)	\(x^{2}+9\)

Differentiate each part:

\(\dfrac{dy}{du}\)	\(=\)	\(\tfrac12 u^{-1/2}\)
\(\dfrac{du}{dx}\)	\(=\)	\(2x\)

Multiply, back-substitute and tidy:

\(\dfrac{dy}{dx}\)	\(=\)	\(\tfrac12 u^{-1/2}\times 2x\)
	\(=\)	\(\dfrac{x}{\sqrt{x^{2}+9}}\)

Evaluate at \(x=4\) (where \(\sqrt{16+9}=5\)):

\(\left.\dfrac{dy}{dx}\right\|_{x=4}\)	\(=\)	\(\dfrac{4}{\sqrt{25}}\)
	\(=\)	\(\dfrac{4}{5}\)

\(\dfrac{dy}{dx}=\dfrac{x}{\sqrt{x^{2}+9}}\), and at \(x=4\) the gradient is \(\dfrac{4}{5}\).

\frac{4}{5}

Common pitfalls

Leaving out the inner derivative. The derivative of \((3x+2)^{4}\) is \(12(3x+2)^{3}\), not \(4(3x+2)^{3}\). You must multiply by \(g'(x)=3\).

Expanding the bracket by mistake. Differentiating the inside is enough; you never expand \((x^{2}-5)^{3}\). Keep the bracket and multiply by its derivative.

Not rewriting roots and reciprocals. Turn \(\sqrt{\ldots}\) into a \(\tfrac12\) power and \(\dfrac{1}{(\ldots)^{n}}\) into a \(-n\) power before the chain rule.

Reducing the power incorrectly. The outer power drops by one: \(u^{4}\to 4u^{3}\), and \(u^{-2}\to -2u^{-3}\). Watch the sign for negative powers.

Frequently asked questions

What is the chain rule?

If \(y=f(u)\) and \(u=g(x)\), then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\): differentiate the outer function and multiply by the derivative of the inner function.

How do you differentiate a bracket raised to a power?

Use \(\dfrac{d}{dx}\big(g(x)\big)^{n}=n\big(g(x)\big)^{n-1}\cdot g'(x)\): reduce the power by one and multiply by the derivative of the inside.

Why is my chain-rule answer wrong by a constant factor?

You have almost certainly left out the inner derivative \(g'(x)\). For \((3x+2)^{4}\) that missing factor is the \(\times 3\).

How do you use the chain rule on a square root?

Rewrite \(\sqrt{g(x)}\) as \((g(x))^{1/2}\), then differentiate: \(\tfrac12(g(x))^{-1/2}\cdot g'(x)\).

What are the inner and outer functions?

The inner function is the expression inside the bracket, \(u=g(x)\); the outer function is what is done to it, here raising to a power, \(y=u^{n}\).

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The Chain Rule

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How to use the chain rule

Common pitfalls

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Theory