Leibniz's Rule/One Variable/Examples/8th Derivative of x^2 sin x
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Example of Use of Leibniz's Rule in One Variable
The $8$th derivative with respect to $x$ of $x^2 \sin x$ is given by:
- $\dfrac {\d^8} {\d x^8} x^2 \sin x = x^2 \sin x - 16 x \cos x - 56 \sin x$
Proof
Leibniz's Rule in One Variable gives:
- $\ds \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$
where $\paren n$ denotes the order of the derivative.
Here we take:
- $\map f x = x^2$
- $\map g x = \sin x$
We note that:
\(\ds \dfrac {\d f} {d x}\) | \(=\) | \(\ds \dfrac \d {d x} x^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 x\) |
\(\ds \dfrac {\d^2 f} {d x^2}\) | \(=\) | \(\ds \dfrac {\d^2} {d x^2} 2 x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2\) |
and for all $n > 2$:
- $\dfrac {\d^n f} {d x^n} = 0$
Hence we need investigate only $\dfrac {\d^n g} {d x^n}$ where $n \in \set {6, 7, 8}$.
Thus:
\(\ds \dfrac {\d^8 g} {d x^8}\) | \(=\) | \(\ds \dfrac {\d^8} {d x^8} \sin x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin x\) |
\(\ds \dfrac {\d^7 g} {d x^7}\) | \(=\) | \(\ds \dfrac {\d^7} {d x^7} \sin x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\cos x\) |
\(\ds \dfrac {\d^6 g} {d x^6}\) | \(=\) | \(\ds \dfrac {\d^6} {d x^6} \sin x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\sin x\) |
Hence:
\(\ds \dfrac {\d^8} {\d x^8} x^2 \sin x\) | \(=\) | \(\ds \sum_{k \mathop = 0}^8 \binom 8 k \map {f^{\paren k} } x \, \map {g^{\paren {8 - k} } } x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom 8 0 x^2 \sin x + \binom 8 1 2 x \paren {-\cos x} + \binom 8 2 2 \paren {-\sin x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 \sin x - 8 \times 2 x \cos x - 28 \times 2 \sin x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 \sin x - 16 x \cos x - 56 \sin x\) |
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Leibniz's Theorem