Leibniz's Rule/One Variable/Second Derivative

From ProofWiki
Jump to navigation Jump to search

Theorem

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:

$\paren {\map f x \, \map g x}'' = \map f x \, \map {g''} x + 2 \map {f'} x \, \map {g'} x + \map {f''} x \, \map g x$


Proof

From Leibniz's Rule:

$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$

where $\paren n$ denotes the order of the derivative.


Setting $n = 2$:

\(\displaystyle \paren {\map f x \, \map g x}''\) \(=\) \(\displaystyle \paren {\map f x \, \map g x}^{\paren 2}\) Definition of Nth Derivative
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^2 \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x\) Leibniz's Rule
\(\displaystyle \) \(=\) \(\displaystyle \binom 2 0 \map {f^{\paren 0} } x \, \map {g^{\paren 2} } x + \binom 2 1 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \binom 2 2 {f^{\paren 2} } x \, \map {g^{\paren 0} } x\)
\(\displaystyle \) \(=\) \(\displaystyle \map {f^{\paren 0} } x \, \map {g^{\paren 2 } } x + 2 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \map {f^{\paren 2} } x \, \map {g^{\paren 0} } x\) Definition of Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \map f x \, \map {g''} x + 2 \map {f'} x \, \map {g'} x + \map {f''} x \, \map g x\) Definition of Nth Derivative

$\blacksquare$


Source of Name

This entry was named for Gottfried Wilhelm von Leibniz.


Sources