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Cambridge International AS & A Level · Topic 1.7
A-Level Maths 9709 / Topic 1.7 / AS & A Level

Outside, then inside.

Roots and reciprocals are just powers in disguise, so one rule handles them all. When the power sits on a bracket, the chain rule adds one extra factor: the derivative of what is inside.

Topic 1.7All differentiation lessons →
The key idea

Write every term as \(kx^{n}\) before differentiating. For a function of a function, \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\): differentiate the outside leaving the inside alone, then multiply by the derivative of the inside. So \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).

xu = 3x − 1y = u⁵ du/dx = 3 dy/du = 5u⁴ dy/dx = (dy/du) × (du/dx) = 5u⁴ × 3 = 15(3x − 1)⁴
Fig. 1: y = (3x − 1)⁵ is a chain of two simple functions; multiply the two rates of change to get dy/dx
Section 01

Any rational power.

The rule \(\dfrac{d}{dx}x^{n} = nx^{n-1}\) works for every rational \(n\): negative, fractional, the lot. The only work is rewriting roots and fractions as powers first, and tidying the answer afterwards.

TermAs a powerDerivative
\(\sqrt{x}\)\(x^{\frac{1}{2}}\)\(\tfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}\)
\(\dfrac{1}{x}\)\(x^{-1}\)\(-x^{-2} = -\dfrac{1}{x^{2}}\)
\(\dfrac{4}{x^{3}}\)\(4x^{-3}\)\(-12x^{-4}\)
\(\dfrac{1}{2x^{2}}\)\(\tfrac{1}{2}x^{-2}\)\(-x^{-3}\)
Worked example Roots and reciprocals
Given \(y = 6\sqrt{x} + \dfrac{8}{x^{2}}\), find \(\dfrac{dy}{dx}\) and the gradient of the curve at \(x = 4\).
  1. Rewrite as powers: \(y = 6x^{\frac{1}{2}} + 8x^{-2}\).
  2. Differentiate: \(6 \times \tfrac{1}{2}x^{-\frac{1}{2}} = 3x^{-\frac{1}{2}}\) and \(8 \times (-2)x^{-3} = -16x^{-3}\).
  3. So \(\dfrac{dy}{dx} = \dfrac{3}{\sqrt{x}} - \dfrac{16}{x^{3}}\).
  4. At \(x = 4\): \(\dfrac{3}{2} - \dfrac{16}{64} = 1.5 - 0.25 = 1.25\).
Answer: \(\dfrac{dy}{dx} = \dfrac{3}{\sqrt{x}} - \dfrac{16}{x^{3}}\); gradient \(\dfrac{5}{4}\) at \(x = 4\).
Worked example Split the fraction first
Differentiate \(y = \dfrac{x^{2} + 3}{\sqrt{x}}\).
  1. You cannot differentiate top and bottom separately. Divide each term by \(x^{\frac{1}{2}}\) instead.
  2. \(y = x^{2 - \frac{1}{2}} + 3x^{-\frac{1}{2}} = x^{\frac{3}{2}} + 3x^{-\frac{1}{2}}\).
  3. \(\dfrac{dy}{dx} = \tfrac{3}{2}x^{\frac{1}{2}} - \tfrac{3}{2}x^{-\frac{3}{2}}\).
Answer: \(\dfrac{dy}{dx} = \tfrac{3}{2}\sqrt{x} - \dfrac{3}{2x^{\frac{3}{2}}}\)
Examiner trap

\(\dfrac{1}{2x^{2}}\) is \(\tfrac{1}{2}x^{-2}\), NOT \(2x^{-2}\): only the \(x^{2}\) moves up with a negative power; the 2 stays in the denominator as \(\tfrac{1}{2}\).

Section 02

The chain rule.

For a bracket raised to a power, let \(u\) be the inside. Then \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\) (Fig. 1). In practice: power down, bracket kept, power reduced by one, times the derivative of the bracket.

Worked example A linear inside
Find the gradient of \(y = (3x - 1)^{5}\) at \(x = 1\).
  1. Let \(u = 3x - 1\), so \(y = u^{5}\), \(\dfrac{dy}{du} = 5u^{4}\) and \(\dfrac{du}{dx} = 3\).
  2. \(\dfrac{dy}{dx} = 5(3x - 1)^{4} \times 3 = 15(3x - 1)^{4}\).
  3. At \(x = 1\): \(15 \times 2^{4} = 15 \times 16 = 240\).
Answer: \(240\)
Worked example A root of a polynomial
Given \(y = \sqrt{2x^{3} + 5}\), find \(\dfrac{dy}{dx}\).
  1. Write \(y = (2x^{3} + 5)^{\frac{1}{2}}\). The inside is \(u = 2x^{3} + 5\), with \(\dfrac{du}{dx} = 6x^{2}\).
  2. Outside: \(\tfrac{1}{2}(2x^{3} + 5)^{-\frac{1}{2}}\).
  3. Multiply: \(\tfrac{1}{2}(2x^{3} + 5)^{-\frac{1}{2}} \times 6x^{2} = 3x^{2}(2x^{3} + 5)^{-\frac{1}{2}}\).
Answer: \(\dfrac{dy}{dx} = \dfrac{3x^{2}}{\sqrt{2x^{3} + 5}}\)
Worked example A reciprocal of a bracket
Differentiate \(y = \dfrac{8}{(2x + 1)^{3}}\) and find the gradient where the curve crosses the \(y\)-axis.
  1. Rewrite: \(y = 8(2x + 1)^{-3}\).
  2. \(\dfrac{dy}{dx} = 8 \times (-3)(2x + 1)^{-4} \times 2 = -48(2x + 1)^{-4}\).
  3. The curve crosses the \(y\)-axis at \(x = 0\): gradient \(= -48 \times 1 = -48\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{48}{(2x + 1)^{4}}\); gradient \(-48\) at \(x = 0\).
Your turn
Differentiate \(y = (4 - x^{2})^{3}\).
Examiner trap

The most common lost mark: forgetting to multiply by the derivative of the inside. \(\dfrac{d}{dx}(3x - 1)^{5}\) is \(15(3x - 1)^{4}\), not \(5(3x - 1)^{4}\). Watch the sign too when the inside is something like \(4 - x^{2}\).

Stage 1 · Learn

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Differentiate \(y = \sqrt{x}\).

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Write \(\dfrac{3}{x^{4}}\) in the form \(kx^{n}\).

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Differentiate \((2x + 1)^{4}\).

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

Section 03

Second derivatives of composites.

To find \(\dfrac{d^{2}y}{dx^{2}}\) for a bracket, apply the chain rule a second time. Keep the answer in factorised form: it is easier to substitute into and to read signs from.

Worked example Chain rule twice
Given \(y = (2x - 3)^{4}\), find \(\dfrac{d^{2}y}{dx^{2}}\) and its value when \(x = 2\).
  1. \(\dfrac{dy}{dx} = 4(2x - 3)^{3} \times 2 = 8(2x - 3)^{3}\).
  2. \(\dfrac{d^{2}y}{dx^{2}} = 8 \times 3(2x - 3)^{2} \times 2 = 48(2x - 3)^{2}\).
  3. At \(x = 2\): \(48 \times 1^{2} = 48\).
Answer: \(\dfrac{d^{2}y}{dx^{2}} = 48(2x - 3)^{2}\); value \(48\).
Examiner trap

Do not expand \((2x - 3)^{4}\) before differentiating unless you have to. It is slow, and a single slip in the binomial expansion costs every later mark.

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A curve has equation \(y = \sqrt{2x^{3} + 5}\). Find the gradient of the curve at the point where \(x = 2\).

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Find the gradient of the curve \(y = \dfrac{6}{(3x - 2)^{2}}\) at the point where \(x = 1\).

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Given \(y = \dfrac{x^{3} - 4}{\sqrt{x}}\), find \(\dfrac{dy}{dx}\) when \(x = 4\).

Exam style+20 XP

Given \(y = (2x - 3)^{4}\), find the value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x = 2\).

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Six original Cambridge-style questions
Rewriting roots and reciprocals as powers, the chain rule for brackets and roots, and a problem on finding where a curve has a given gradient.