Leibniz's Rule/One Variable
Theorem
Let $f$ and $g$ be real functions defined on the open interval $I$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are $n$ times differentiable.
Then:
- $\ds \paren {\map f x \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n - k} } } x$
where $\paren n$ denotes the order of the derivative.
Proof
Proof by induction:
Basis for the Induction
Let $n = 1$.
From Product Rule for Derivatives:
- $\paren {\map f x \map g x}' = \paren {\map f x \map {g'} x} + \paren {\map {f'} x \map g x}$
Likewise:
- $\ds \sum_{k \mathop = 0}^1 \binom 1 k \map {f^{\paren k} } x \map {g^{\paren {1 - k} } } x = \binom 1 0 \map f x \map {g^{\paren {1 - 0} } } x + \binom 1 1 \map {f'} x \map {g^{\paren {1 - 1} } } x = \map f x \map {g'} x + \map {f'} x \map g x$
This is our basis for the induction.
Induction Hypothesis
Let $n \in \N$ be fixed.
We assume the inductive hypothesis:
- $\ds \paren {\map f x \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n - k} } } x$
We need to show that:
- $\ds \paren {\map f x \map g x}^{\paren {n + 1} } = \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x$
Induction Step
By our inductive hypothesis:
\(\ds \paren {\map f x \map g x}^{\paren {n + 1} }\) | \(=\) | \(\ds \paren {\sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n - k} } } x}'\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {\paren {\map {f^{\paren k} } x \map {g^{\paren {n - k} } } x}'}\) | Applications of Linear Combination of Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {\map {f^{\paren {k + 1} } } x \map {g^{\paren {n - k} } } x + \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x}\) | from the base case | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren {k + 1} } } x \map {g^{\paren {n - k} } } x + \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x\) | splitting the summation |
Subsequently, we separate the $k = 0$ case from the second summation.
For the first summation, we separate the case $k = n$ and then shift the indices up by $1$.
These manipulations give us the following:
\(\ds \paren {\map f x \map g x}^{\paren {n + 1} }\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x + \sum_{k \mathop = 1}^n \binom n {k - 1} \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x + \binom n 0 \map f x \map {g^{\paren {n + 1} } } x + \binom n n \map {f^{\paren {n + 1} } } x \map g x\) |
By Pascal's Rule, we finally obtain:
\(\ds \paren {\map f x \map g x}^{\paren {n + 1} }\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \binom {n + 1} k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x + \binom n 0 \map f x \map {g^{\paren {n + 1} } } x + \binom n n \map {f^{\paren {n + 1} } } x \map g x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \binom {n + 1} k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x + \binom {n + 1} 0 \map f x \map {g^{\paren {n + 1} } } x + \binom {n + 1} {n + 1} \map {f^{\paren {n + 1} } } x \map g x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k \map {f^{\paren k} } x \map {g^{\paren {n + 1 - k} } } x\) |
The result follows by the Principle of Mathematical Induction.
$\blacksquare$
Also known as
Leibniz's Rule is also known as Leibniz's theorem or Leibniz theorem.
Special Cases
Second Derivative
Let $f$ and $g$ be real functions defined on the open interval $I$.
Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are twice differentiable.
Then:
- $\paren {\map f x \map g x}'' = \map f x \map {g''} x + 2 \map {f'} x \map {g'} x + \map {f''} x \map g x$
Third Derivative
Let $f$ and $g$ be real functions defined on the open interval $I$.
Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are thrice differentiable.
Then:
- $\paren {\map f x \map g x}''' = \map f x \map {g'''} x + 3 \map {f'} x \map {g''} x + 3 \map {f''} x \map {g'} x + \map {f'''} x \map g x$
Examples
$8$th Derivative of $x^2 \sin x$
The $8$th derivative with respect to $x$ of $x^2 \sin x$ is given by:
- $\dfrac {\d^8} {\d x^8} x^2 \sin x = x^2 \sin x - 16 x \cos x - 56 \sin x$
Source of Name
This entry was named for Gottfried Wilhelm von Leibniz.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Leibnitz's Rule for Higher Derivatives of Products: $13.46$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 10.11 \ (6)$
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Leibniz's theorem
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Leibniz theorem
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Leibniz theorem
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Leibniz's Theorem