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Higher derivatives and Maclaurin series

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

A Maclaurin series writes a function as an infinite polynomial in x, built from its derivatives at x = 0. You derive the standard series for e^x, sin x, cos x and ln(1 + x), and the method works for any function that is differentiable enough at 0.

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Higher derivatives

f'''(x) or d^3y/dx³ is the third derivative; fⁿ(x) is the nth. Differentiate one step at a time, using the product, quotient and chain rules, and simplify as you go.

For y = e^2x sin x, or an implicit relation, write each derivative on its own line.

The Maclaurin series

f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... + f^r(0)x^r/r! + ...

Quote this general form before you substitute numbers: the scheme of work says that without it, method marks may not be awarded. Remember the factorials.

Standard series

e^x = 1 + x + x²/2! + x³/3! + ..., valid for all x
sin x = x - x³/3! + x⁵/5! - ..., valid for all x
cos x = 1 - x²/2! + x⁴/4! - ..., valid for all x
ln(1 + x) = x - x²/2 + x³/3 - ..., valid for -1 < x ≤ 1

These are in the formula booklet; you may still be asked to derive them.

When there is no series

A Maclaurin series needs f(0), f'(0), f''(0), ... to exist. ln x has none, because ln 0 is undefined; that is why the standard series is for ln(1 + x). Truncated series are good approximations for x close to 0.

Worked example

Find the Maclaurin series for ln(1 + x) up to the term in x⁴.

f(x) = ln(1 + x), f(0) = 0.
f'(x) = (1 + x)^-1, f'(0) = 1; f''(x) = -(1 + x)^-2, f''(0) = -1.
f'''(x) = 2(1 + x)^-3, f'''(0) = 2; f''''(x) = -6(1 + x)^-4, f''''(0) = -6.
f(x) = 0 + x - x²/2! + 2x³/3! - 6x⁴/4!.

Answer: ln(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ...

Key idea: f(x) = Σ f^r(0)x^r/r!. Find the derivatives, evaluate at 0, quote the formula, then substitute.

Check yourself

f(x) = x² e^x. Find f'''(0).
Show answer
6. x² e^x = x² + x³ + x⁴/2 + ..., so the coefficient of x³ is 1 = f'''(0)/3!. f'''(0) = 6.
Find the coefficient of x⁵ in the Maclaurin series for sin x. Give your answer as a fraction in its simplest form.
Show answer
1/120. sin x = x - x³/3! + x⁵/5! - ..., so the coefficient is 1/5! = 1/120.
For which values of x is the Maclaurin series of ln(1 + x) valid?
- -1 < x ≤ 1
- All real x
- -1 ≤ x ≤ 1
- x > -1
Show answer
-1 < x ≤ 1. The series converges for -1 < x ≤ 1. At x = -1, ln 0 is undefined.
Use differentiation to find the coefficient of x² in the Maclaurin series for √(1 + x). Give your answer as a fraction in its simplest form.
Show answer
-1/8. f''(x) = -¼(1 + x)^-3/2, so f''(0) = -1/4 and the coefficient is (-1/4)/2! = -1/8. This agrees with the binomial expansion.
Use the Maclaurin series for e^x up to the term in x³ to estimate e⁰.1, to 5 decimal places.
Show answer
1.10517. 1 + 0.1 + 0.01/2 + 0.001/6 = 1.105167, which is 1.10517 to 5 decimal places.
Why does ln x have no Maclaurin series?
- ln x is not continuous for x > 0
- The derivatives of ln x are all zero
- ln x has no derivatives
- ln x and its derivatives are undefined at x = 0
Show answer
ln x and its derivatives are undefined at x = 0. The series needs f(0), f'(0), ...; ln 0 and 1/0 are undefined.
f(x) = tan x has f'''(0) = 2. Find the coefficient of x³ in its Maclaurin series. Give your answer as a fraction in its simplest form.
Show answer
1/3. Coefficient = f'''(0)/3! = 2/6 = 1/3. So tan x = x + x³/3 + ....
f(x) = e^{sin x}. Find f''(0).
Show answer
1. f' = cos x e^{sin x}, f'' = (cos² x - sin x)e^{sin x}. At x = 0: f''(0) = (1 - 0) × 1 = 1.

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Last updated 8 November 2026. Spotted a mistake? Tell us.