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I don't fully understand the rule of Leibniz and I'm trying to find the 12th order derivative of: $x\cos \left(x\right)$

How do I find this using the Leibniz rule?

AsukaMinato
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Stanko
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4 Answers4

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The answer in summation notation is $$\sum_{k=0}^{12}{\binom{12}{k}\frac{d^k}{dx^k}x\frac{d^{n-k}}{dx^{n-k}}{\cos{x}}}$$ This looks formidable, but most of the terms are 0. If we write it out we get $$x\frac{d^{12}}{dx^{12}}\cos x+12\frac{d^{11}}{dx^{11}}\cos x$$ or $$x\cos{x}+12\sin{x}$$ This is because derivatives of $x$ beyond the first are 0. To explain the trigonometric terms, note that the cosine function is its own fourth derivative.

Matt Samuel
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  • It might be helpful to write $\sum_{k=0}^{12} {12 \choose k} \frac{d^k}{dx^k} x \frac{d^{n-k}}{dx^{n-k}} \cos(x) = x \frac{d^{12}}{dx^{12}} \cos(x) + 12 \frac{d^{11}}{dx^{11}} \cos(x)$. – Ian Dec 26 '14 at 16:39
  • @Ian thanks for the suggestion, added. – Matt Samuel Dec 26 '14 at 16:43
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The Leibniz rule says that $$(fg)'=f'g + fg'$$ So, in your case, we have $f(x)=\cos(x)$ and $g(x)=x$. I assume you have the identities handy that $f'(x)=-\sin(x)$ and $g'(x)=1$. So, plugging this into the above gives the first derivative as $$-x\sin(x)+\cos(x)$$ What we should notice is that, the next time we differentiate, we have $-x\sin(x)$ being the product of two terms - but the derivative of $-x$ is $-1$ and the derivative of $\sin(x)$ is $\cos(x)$, which, plugging that into Leibniz rule yields the derivative of $-x\sin(x)$ to be $$-x\cos(x)-\sin(x)$$ and, adding that to the derivative of $\cos(x)$ to get the second derivative of your function yields: $$-x\cos(x)-2\sin(x).$$ You can proceed thusly, since now we have another $x\cos(x)$ term. You will probably notice a pattern as you compute the 12 derivatives, but it's not prohibitively difficult to do by hand.

Milo Brandt
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  • OP is talking about the general Leibniz rule http://en.wikipedia.org/wiki/General_Leibniz_rule , not the product rule. – Ganesh Dec 26 '14 at 19:06
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For brevity, let's write $c:=\cos x$ and $s:=\sin x$, and write the $n$th derivative of $y_0:=xc$ with respect to $x$ as $y_n$ ($n=1,2,...$). It's pretty easy to spot the pattern of the even-numbered derivatives as $$y_{2n}=(-1)^n(xc+2ns)\;\;(n=0,1,...),$$ and this can be proved straightforwardly by induction.

John Bentin
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By observation I found

$$\frac{d^n}{dx^n}x\cos x = \begin{cases} (-1)^{n/2}n \sin x + (-1)^{n/2}x\cos x, & \text{if $n$ is even} \\ (-1)^{(n-1)/2}n \cos x + (-1)^{(n+1)/2}x\sin x, & \text{if $n$ is odd} \end{cases}$$

which could be proved by induction and product rule. Plugging in we get

$$\frac{d^{12}}{dx^{12}}x\cos x = 12\sin x + x\cos x$$

David Etler
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