Write the Maclaurin series of e^x up to the term in x^3. [1 mark]

Use series to evaluate _x→ 01 - cos xx^2. [2 marks]

SEAB Original-style practice: Given that y = ln(1 + sin x), show that dydx = cos x1 + sin x, and hence find the Maclaurin series of y up to and including the term in x^3.

Differentiate: dydx = 11 + sin x·cos x = cos x1 + sin x. It is cleaner to multiply through: (1 + sin x)dydx = cos x. Differentiate again (product rule on the left): $cos x\,dydx + (1 + sin x)d^2ydx^2 = -sin x.$ Differentiate once more: $-sin x\,dydx + cos x\,d^2ydx^2 + cos x\,d^2ydx^2 + (1 + sin x)d^3ydx^3 = -cos x.$ Now evaluate at x = 0: y = ln 1 = 0; dydx = 11 = 1; from the second equation 11 + 1· y'' = 0 y'' = -1; from the third 0 + 2(−1) + y''' = -1 y''' = 1. Maclaurin: y = 0 + 1· x + -12!x^2 + 13!x^3 + = x - x^22 + x^36 + Markers reward the first derivative, repeated implicit differentiation, the values of the derivatives at 0, and the correct series with factorials.

SEAB Original-style practice: Use Maclaurin series to evaluate _x → 0 (x - sin x)/(x^3).

Use the standard expansion sin x = x - x^36 + x^5120 -. Then $x - sin x = x - (x - (x^3)/(6) + (x^5)/(120) - ) = (x^3)/(6) - (x^5)/(120) + .$ Divide by x^3: $(x - sin x)/(x^3) = (1)/(6) - (x^2)/(120) + .$ As x → 0 the remaining terms vanish, so the limit is 16. Markers reward substituting the series for sin x, the leading non-zero term x^36, dividing by x^3, and the limit 16.

SingaporeFurther MathsSyllabus dot point

How do we derive a Maclaurin series, including by repeated implicit differentiation, and use it for limits and approximations?

Derive Maclaurin series including by repeated implicit differentiation and use series to evaluate limits and approximations

A focused answer to the H2 Further Mathematics outcome on Maclaurin series. The general formula, deriving series by repeated and implicit differentiation, combining standard expansions, and using series to evaluate limits and small-value approximations.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to derive Maclaurin series, including the harder case where the function is given implicitly and you must differentiate the defining equation repeatedly, and to use series for two applications: evaluating limits of the indeterminate $\tfrac{0}{0}$ type and approximating function values. At Further level the emphasis is on the implicit-differentiation derivation and on series as a tool for limits.

The answer

The Maclaurin formula

The Maclaurin series expands a function as a power series about $x = 0$ :

\mathrm{f}(x) = \mathrm{f}(0) + \mathrm{f}'(0)\,x + \frac{\mathrm{f}''(0)}{2!}\,x^2 + \frac{\mathrm{f}'''(0)}{3!}\,x^3 + \cdots = \sum_{r=0}^{\infty} \frac{\mathrm{f}^{(r)}(0)}{r!}\,x^r.

Deriving a series by repeated differentiation

When derivatives are easy, differentiate $\mathrm{f}$ repeatedly, evaluate each derivative at $0$ , and substitute. For a function defined implicitly (for example $y = \ln(1 + \sin x)$ ), it is usually neater to clear the denominator to get a relation such as $(1 + \sin x)y' = \cos x$ , then differentiate that relation repeatedly with the product rule, evaluating at $x = 0$ at each stage to generate $y', y'', y''', \dots$ in turn.

The standard series

The expansions to know:

\mathrm{e}^x = 1 + x + \frac{x^2}{2!} + \cdots, \quad \sin x = x - \frac{x^3}{3!} + \cdots, \quad \cos x = 1 - \frac{x^2}{2!} + \cdots,

\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\ (-1 < x \leq 1), \quad (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots\ (|x| < 1).

Combining series

Build new series by substitution (for example $x \to 2x$ ), multiplication, or term-by-term differentiation and integration. This is faster than repeated differentiation when a standard series applies.

Series for limits

A limit of the form $\dfrac{0}{0}$ as $x \to 0$ is found by replacing numerator and denominator with their series and cancelling the lowest powers of $x$ . The surviving constant term is the limit. This is a clean alternative to repeated L'Hopital differentiation.

Worked example

Find the Maclaurin series of $y = \mathrm{e}^{\sin x}$ up to the term in $x^3$ .

Step 1: Set up a relation by differentiating

$\dfrac{dy}{dx} = \mathrm{e}^{\sin x}\cos x = y\cos x$ , so $y' = y\cos x$ .

Step 2: Differentiate the relation again

By the product rule, $y'' = y'\cos x - y\sin x$ .

Step 3: Differentiate once more

y''' = y''\cos x - y'\sin x - y'\sin x - y\cos x = y''\cos x - 2y'\sin x - y\cos x.

Step 4: Evaluate everything at x = 0

At $x = 0$ : $\sin 0 = 0$ , $\cos 0 = 1$ , and $y(0) = \mathrm{e}^0 = 1$ .

$y'(0) = y(0)\cos 0 = 1$ . $y''(0) = y'(0)(1) - y(0)(0) = 1$ . $y'''(0) = y''(0)(1) - 2y'(0)(0) - y(0)(1) = 1 - 0 - 1 = 0$ .

Step 5: Assemble the series

y = 1 + 1\cdot x + \frac{1}{2!}x^2 + \frac{0}{3!}x^3 + \cdots = 1 + x + \frac{x^2}{2} + \cdots

Examples in context

Example 1. Replacing L'Hopital's rule. A limit like $\displaystyle\lim_{x\to 0}\dfrac{1 - \cos x}{x^2}$ is read straight off the series $1 - \cos x = \tfrac{x^2}{2} - \cdots$ , giving $\tfrac{1}{2}$ in one line where repeated differentiation would be slower.

Example 2. Linearising for physics. Approximating $\mathrm{e}^{\sin x} \approx 1 + x$ for small $x$ , or $\cos x \approx 1 - \tfrac{x^2}{2}$ , is exactly the small-oscillation approximation used to linearise pendulum and circuit equations before solving them.

Try this

Q1. Write the Maclaurin series of $\mathrm{e}^x$ up to the term in $x^3$ . [1 mark]

Cue. $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$ .

Q2. Use series to evaluate $\displaystyle\lim_{x\to 0}\dfrac{1 - \cos x}{x^2}$ . [2 marks]

Cue. $1 - \cos x = \tfrac{x^2}{2} - \cdots$ , so the ratio tends to $\tfrac{1}{2}$ .

Q3. If $y = (1 + x)^{1/2}$ , give the first three terms of its Maclaurin series. [2 marks]

Cue. Binomial with $n = \tfrac{1}{2}$ : $1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \cdots$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original7 marksGiven that

y = \ln(1 + \sin x)

, show that

\dfrac{dy}{dx} = \dfrac{\cos x}{1 + \sin x}

, and hence find the Maclaurin series of

y

up to and including the term in

x^3

Show worked answer →

Differentiate: $\dfrac{dy}{dx} = \dfrac{1}{1 + \sin x}\cdot\cos x = \dfrac{\cos x}{1 + \sin x}$ .

It is cleaner to multiply through: $(1 + \sin x)\dfrac{dy}{dx} = \cos x$ . Differentiate again (product rule on the left):

\cos x\,\dfrac{dy}{dx} + (1 + \sin x)\dfrac{d^2y}{dx^2} = -\sin x.

Differentiate once more:

-\sin x\,\dfrac{dy}{dx} + \cos x\,\dfrac{d^2y}{dx^2} + \cos x\,\dfrac{d^2y}{dx^2} + (1 + \sin x)\dfrac{d^3y}{dx^3} = -\cos x.

Now evaluate at

x = 0

y = \ln 1 = 0

;

\dfrac{dy}{dx} = \dfrac{1}{1} = 1

; from the second equation

1\cdot1 + 1\cdot y'' = 0 \Rightarrow y'' = -1

; from the third

0 + 2(−1) + y''' = -1 \Rightarrow y''' = 1

Maclaurin: $y = 0 + 1\cdot x + \dfrac{-1}{2!}x^2 + \dfrac{1}{3!}x^3 + \cdots = x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$

Markers reward the first derivative, repeated implicit differentiation, the values of the derivatives at $0$ , and the correct series with factorials.

Original4 marksUse Maclaurin series to evaluate

\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}

Show worked answer →

Use the standard expansion $\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$ . Then

x - \sin x = x - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) = \frac{x^3}{6} - \frac{x^5}{120} + \cdots.

Divide by

x^3

\frac{x - \sin x}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots.

x \to 0

the remaining terms vanish, so the limit is

\dfrac{1}{6}

Markers reward substituting the series for $\sin x$ , the leading non-zero term $\tfrac{x^3}{6}$ , dividing by $x^3$ , and the limit $\tfrac{1}{6}$ .