Leibniz's Rule/One Variable

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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:

$\paren {\map f x \map g x}' = \map f x \map {g'} x + \map {f'} x \map g x$

Likewise:

\(\ds \sum_{k \mathop = 0}^1 \binom 1 k \map {f^{\paren k} } x \map {g^{\paren {1 - k} } } x\) \(=\) \(\ds \binom 1 0 \map {f^{\paren 0} } x \map {g^{\paren {1 - 0} } } x + \binom 1 1 \map {f^{\paren 1} } x \map {g^{\paren {1 - 1} } } x\)
\(\ds \) \(=\) \(\ds \map f x \map {g'} x + \map {f'} x \map g x\)


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 presented as

Leibniz's Rule in One Variable can often be seen presented in this format:

\(\ds \map {\dfrac {\d^n} {\d x^n} } {f g}\) \(=\) \(\ds \sum_{k \mathop = 0}^n \dbinom n k \dfrac {\d^k f} {\d x^k} \dfrac {\d^{n - k} g} {\d x^{n - k} }\)
\(\ds \) \(=\) \(\ds \dfrac {\d^n f} {\d x^n} g + \dfrac {\d^{n - 1} f} {\d x^{n - 1} } \dfrac {\d g} {\d x} + \dfrac {\d^{n - 2} f} {\d x^{n - 2} } \dfrac {\d^2 g} {\d x^2} + \cdots + \dfrac {\d^{n - k} f} {\d x^{n - k} } \dfrac {\d^k g} {\d^k x} + \cdots + \dfrac {\d f} {\d x} \dfrac {\d^{n - 1} g} {\d^{n - 1} x} + f \dfrac {\d^n g} {\d x^n}\)


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

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