A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther CalculusMethods in CalculusLeibnitz's Theorem

Leibnitz's Theorem (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 6 January 2026

Leibnitz's theorem

What is Leibnitz's theorem?

Leibnitz's theorem states that the n^th derivative of a product of functions, $y = f (x) g (x)$ , is given by
- $\begin{array}{rcl} \frac{d^{n} y}{d x^{n}} & = & (\begin{matrix} n \\ 0 \end{matrix}) f^{(0)} (x) g^{(n)} (x) + (\begin{matrix} n \\ 1 \end{matrix}) f^{(1)} (x) g^{(n - 1)} (x) + (\begin{matrix} n \\ 2 \end{matrix}) f^{(2)} (x) g^{(n - 2)} (x) + . . . + (\begin{matrix} n \\ r \end{matrix}) f^{(n - r)} (x) g^{(r)} (x) + . . . \\ . . . + (\begin{matrix} n \\ n - 1 \end{matrix}) f^{(n - 1)} (x) g^{(1)} (x) + (\begin{matrix} n \\ n \end{matrix}) f^{(n)} (x) g^{(0)} (x) \end{array}$
- where
  - $f^{(0)} (x) = f (x)$
  - $f^{(r)} (x)$ means the r^th derivative of $f (x)$
  - and $(\begin{matrix} n \\ r \end{matrix})$ is the binomial coefficient $\frac{n!}{r! (n - r)!}$
It allows you to find $\frac{d^{n} y}{d x^{n}}$ from $y = f (x) g (x)$ directly
- without having to differentiate n times!
In the case when $n = 1$ you get the product rule
- i.e. if $y = f (x) g (x)$ then
  - $\frac{d y}{d x} = (\begin{matrix} 1 \\ 0 \end{matrix}) f^{(0)} (x) g^{(1)} (x) + (\begin{matrix} 1 \\ 1 \end{matrix}) f^{(1)} (x) g^{(0)} (x) = f (x) g' (x) + f' (x) g (x)$

Examiner Tips and Tricks

You need to learn Leibnitz's theorem as it is not given in the formulae booklet!

How do I use Leibnitz's theorem?

To use Leibnitz's theorem, it helps to write a table of derivatives then match opposite ends
- e.g. if $y = e^{2 x} \sin x$ find $\frac{d^{4} y}{d x^{4}}$
- Calculate the derivatives
  $r$
  $f^{(r)} (x)$
  $g^{(r)} (x)$
  0
  $e^{2 x}$
  $\sin x$
  1
  $2 e^{2 x}$
  $\cos x$
  2
  $4 e^{2 x}$
  $- \sin x$
  3
  $8 e^{2 x}$
  $- \cos x$
  4
  $16 e^{2 x}$
  $\sin x$
- Match opposite ends (the first $f$ with the last $g$ , second $f$ with second-to-last $g$ , etc)
  $r$
  $f^{(r)} (x)$
  $g^{(r)} (x)$
  0
  $e^{2 x}$
  $\sin x$
  1
  $2 e^{2 x}$
  $\cos x$
  2
  $4 e^{2 x}$
  $- \sin x$
  3
  $8 e^{2 x}$
  $- \cos x$
  4
  $16 e^{2 x}$
  $\sin x$
- Find the relevant binomial coefficients $(\begin{matrix} 4 \\ 0 \end{matrix})$ , $(\begin{matrix} 4 \\ 1 \end{matrix})$ , $(\begin{matrix} 4 \\ 2 \end{matrix})$ , $(\begin{matrix} 4 \\ 3 \end{matrix})$ and $(\begin{matrix} 4 \\ 4 \end{matrix})$
  - 1, 4, 6, 4, 1
- Substitute into Leibnitz's theorem
  - $\frac{d^{4} y}{d x^{4}} = 1 \times e^{2 x} \sin x + 4 \times 2 e^{2 x} (- \cos x) + 6 \times 4 e^{2 x} (- \sin x) + 4 \times 8 e^{2 x} \cos x + 1 \times 16 e^{2 x} \sin x$
- Simplify and collect like terms
  - $\begin{array}{rcl} \frac{d^{4} y}{d x^{4}} & = & e^{2 x} \sin x - 8 e^{2 x} \cos x - 24 e^{2 x} \sin x + 32 e^{2 x} \cos x + 16 e^{2 x} \sin x \\ = & - 7 e^{2 x} \sin x + 24 e^{2 x} \cos x \end{array}$

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$e^{2 x}$	$\sin x$
1	$2 e^{2 x}$	$\cos x$
2	$4 e^{2 x}$	$- \sin x$
3	$8 e^{2 x}$	$- \cos x$
4	$16 e^{2 x}$	$\sin x$

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$e^{2 x}$	$\sin x$
1	$2 e^{2 x}$	$\cos x$
2	$4 e^{2 x}$	$- \sin x$
3	$8 e^{2 x}$	$- \cos x$
4	$16 e^{2 x}$	$\sin x$

Examiner Tips and Tricks

A useful check, before simplifying, is that the n^th derivative should have n+1 terms (e.g. $\frac{d^{4} y}{d x^{4}}$ should have 5 terms).

How do I prove general results using Leibnitz's theorem?

You can use Leibnitz's theorem to prove general results like the following:
- If $y = x^{2} e^{2 x}$ , prove that $\frac{d^{n} y}{d x^{n}} = 2^{n - 2} e^{2 x} (4 x^{2} + 4 n x + n^{2} - n)$ for $n \in ℕ$
You need to know the following properties of the binomial coefficients
- $(\begin{matrix} n \\ 0 \end{matrix}) = 1$ , $(\begin{matrix} n \\ 1 \end{matrix}) = n$ and by symmetry $(\begin{matrix} n \\ n \end{matrix}) = 1$ , $(\begin{matrix} n \\ n - 1 \end{matrix}) = n$
- Higher-order coefficients can be simplified
  - $(\begin{matrix} n \\ 2 \end{matrix}) = \frac{n!}{2! (n - 2)!} = \frac{n (n - 1) (n - 2)!}{2 \times 1 \times (n - 2)!} = \frac{1}{2} n (n - 1)$
  - $(\begin{matrix} n \\ 3 \end{matrix}) = \frac{n!}{3! (n - 3)!} = \frac{n (n - 1) (n - 2) (n - 3)!}{3 \times 2 \times 1 \times (n - 3)!} = \frac{1}{6} n (n - 1) (n - 2)$
  - etc.
You need to be able to spot patterns in the derivatives
- See the worked example below

Worked Example

Using Leibnitz's theorem,

(a) find $a$ and $b$ such that $\frac{d^{6} y}{d x^{6}} = a + b \ln x$ where $y = x^{6} \ln x$ .

(b) prove that if $y = x^{2} e^{2 x}$ then $\frac{d^{n} y}{d x^{n}} = 2^{n - 2} e^{2 x} (4 x^{2} + 4 n x + n^{2} - n)$ for $n \in ℕ$ .

Answer:

(a)

Let $f (x) = x^{6}$ and $g (x) = \ln x$

Work out a table of derivatives

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$x^{6}$	$\ln x$
1	$6 x^{5}$	$x^{- 1}$
2	$30 x^{4}$	$- x^{- 2}$
3	$120 x^{3}$	$2 x^{- 3}$
4	$360 x^{2}$	$- 6 x^{- 4}$
5	$720 x$	$24 x^{- 5}$
6	$720$	$120 x^{- 6}$

Match opposite ends (the first $f$ with the last $g$ , second $f$ with second-to-last $g$ , etc)

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$x^{6}$	$\ln x$
1	$6 x^{5}$	$x^{- 1}$
2	$30 x^{4}$	$- x^{- 2}$
3	$120 x^{3}$	$2 x^{- 3}$
4	$360 x^{2}$	$- 6 x^{- 4}$
5	$720 x$	$24 x^{- 5}$
6	$720$	$- 120 x^{- 6}$

Find the binomial coefficients $(\begin{matrix} 6 \\ 0 \end{matrix})$ , $(\begin{matrix} 6 \\ 1 \end{matrix})$ , $(\begin{matrix} 6 \\ 2 \end{matrix})$ , ..., $(\begin{matrix} 6 \\ 6 \end{matrix})$

1, 6, 15, 20, 15, 6, 1

Leibnitz's theorem for $n = 6$ is

$\begin{array}{rcl} \frac{d^{6} y}{d x^{6}} & = & (\begin{matrix} 6 \\ 0 \end{matrix}) f^{(0)} (x) g^{(6)} (x) + (\begin{matrix} 6 \\ 1 \end{matrix}) f^{(1)} (x) g^{(5)} (x) + (\begin{matrix} 6 \\ 2 \end{matrix}) f^{(2)} (x) g^{(4)} (x) + . . . \\ . . . + (\begin{matrix} 6 \\ 5 \end{matrix}) f^{(5)} (x) g^{(1)} (x) + (\begin{matrix} 6 \\ 6 \end{matrix}) f^{(6)} (x) g^{(0)} (x) \end{array}$

Substitute the above into Leibnitz's theorem

$\begin{array}{l} \frac{d^{6} y}{d x^{6}} = 1 \times x^{6} \times (- 120 x^{- 6}) + 6 \times 6 x^{5} \times 24 x^{- 5} + 15 \times 30 x^{4} \times (- 6 x^{- 4}) + . . . \\ 20 \times 120 x^{3} \times 2 x^{- 3} + 15 \times 360 x^{2} \times (- x^{- 2}) + 6 \times 720 x \times x^{- 1} + 1 \times 720 \times \ln x \end{array}$

Simplify each term

$\frac{d^{6} y}{d x^{6}} = - 120 + 864 - 2700 + 4800 - 5400 + 4320 + 720 \ln x$

Collect like terms

$\frac{d^{6} y}{d x^{6}} = 1764 + 720 \ln x$

This is now in the form given in the question, $\frac{d^{6} y}{d x^{6}} = a + b \ln x$ , so state $a$ and $b$

$a = 1764 and b = 720$

(b)

Start by writing a table of derivatives for $f (x) = x^{2}$ and $g (x) = e^{2 x}$

Notice that $x^{2}$ eventually differentiates to zero
Add in a few rows at the bottom for n-2, n-1, n

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$x^{2}$	$e^{2 x}$
1	$2 x$	$2 e^{2 x}$
2	$2$	$4 e^{2 x}$
3	$0$	$8 e^{2 x}$
4	$0$	$16 e^{2 x}$
...	...	...
n-2	$0$
n-1	$0$
n	$0$

Spot a general rule / pattern in the $g$ derivatives to find $g^{(n)} (x)$

The coefficients of $e^{2 x}$ are powers of 2

$g^{(n)} (x) = 2^{n} e^{2 x}$

Use this rule to backfill $g^{(n - 1)} (x)$ and $g^{(n - 2)} (x)$

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$x^{2}$	$e^{2 x}$
1	$2 x$	$2 e^{2 x}$
2	$2$	$4 e^{2 x}$
3	$0$	$8 e^{2 x}$
4	$0$	$16 e^{2 x}$
...	...	...
n-2	$0$	$2^{n - 2} e^{2 x}$
n-1	$0$	$2^{n - 1} e^{2 x}$
n	$0$	$2^{n} e^{2 x}$

Match opposite ends (the first $f$ with the last $g$ , second $f$ with second-to-last $g$ , etc)

This makes all terms after the third term equal to zero

$r$	$f^{(r)} (x)$	$g^{(r)} (x)$
0	$x^{2}$	$e^{2 x}$
1	$2 x$	$2 e^{2 x}$
2	$2$	$4 e^{2 x}$
3	$0$	$8 e^{2 x}$
4	$0$	$16 e^{2 x}$
...	...	...
n-2	$0$	$2^{n - 2} e^{2 x}$
n-1	$0$	$2^{n - 1} e^{2 x}$
n	$0$	$2^{n} e^{2 x}$

Find and simplify the first three binomial coefficients in terms of $n$

$\begin{array}{rcl} (\begin{matrix} n \\ 0 \end{matrix}) & = & 1 \\ (\begin{matrix} n \\ 1 \end{matrix}) & = & n \\ (\begin{matrix} n \\ 2 \end{matrix}) & = & \frac{n}{2! (n - 2)!} = \frac{n (n - 1) (n - 2)!}{2 \times 1 \times (n - 2)!} = \frac{1}{2} n (n - 1) \end{array}$

Write down Leibnitz's theorem in terms of $n$

$\begin{array}{rcl} \frac{d^{n} y}{d x^{n}} & = & (\begin{matrix} n \\ 0 \end{matrix}) f^{(0)} (x) g^{(n)} (x) + (\begin{matrix} n \\ 1 \end{matrix}) f^{(1)} (x) g^{(n - 1)} (x) + (\begin{matrix} n \\ 2 \end{matrix}) f^{(2)} (x) g^{(n - 2)} (x) + . . . \\ . . . + (\begin{matrix} n \\ n - 1 \end{matrix}) f^{(n - 1)} (x) g^{(1)} (x) + (\begin{matrix} n \\ n \end{matrix}) f^{(n)} (x) g^{(0)} (x) \end{array}$

Substitute the above working into Leibnitz's theorem

$\frac{d^{n} y}{d x^{n}} = 1 \times x^{2} \times 2^{n} e^{2 x} + n \times 2 x \times 2^{n - 1} e^{2 x} + \frac{1}{2} n (n - 1) \times 2 \times 2^{n - 2} e^{2 x} + 0 + 0 + 0 + . . .$

Simplify each term

$\frac{d^{n} y}{d x^{n}} = 2^{n} e^{2 x} x^{2} + 2^{n} e^{2 x} n x + 2^{n - 2} e^{2 x} (n^{2} - n)$

By comparing to the answer given, factorise out a $2^{n - 2}$ and an $e^{2 x}$

$\frac{d^{n} y}{d x^{n}} = 2^{n - 2} e^{2 x} (2^{2} x^{2} + 2^{2} n x + n^{2} - n)$

This simplifies to the correct answer

$\frac{d^{n} y}{d x^{n}} = 2^{n - 2} e^{2 x} (4 x^{2} + 4 n x + n^{2} - n) for n \in ℤ$

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