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Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$.

The answer is too short for me to understand, and the answer is

$- 25 \cdot 24 \cdot 8^{23}$

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    $\begingroup$ HINT Take the derivative 25 times, then plug in "0" $\endgroup$ – Don Larynx Dec 23 '13 at 12:32
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    $\begingroup$ @DonLarynx : That would not be a better idea and i believe OP knows the definition of what it is written and what you have said is nothing more than a definition $\endgroup$ – user87543 Dec 23 '13 at 12:34
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    $\begingroup$ What's the Maclaurin series for your function? $\endgroup$ – David Mitra Dec 23 '13 at 12:34
  • $\begingroup$ @DavidMitra : I do not quite understand your idea... you need that derivative to write maclaurian series right? I am missing something I guess... $\endgroup$ – user87543 Dec 23 '13 at 12:38
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    $\begingroup$ @PraphullaKoushik You can find the Maclauren series of $f(x)=x^2\sin x$ by multiplying the Maclauren series for $\sin$ by $x^2$. $\endgroup$ – David Mitra Dec 23 '13 at 12:40
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$$f(x) = x^2\sin x$$ $$ = x^2 \left(x - \frac{x^3}{3!} +...- \frac{x^{23}}{23!} +...\right)$$ $$ = -\frac{x^{25}}{23!} +...$$ $$f^{(25)}(x) = -\frac{25!}{23!} + ...$$ $$f^{(25)}(0) = -600$$

WolframAlpha verifies that the $8^{23}$ should not be there.

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use the leibnitz rule to find the nth derivative of the function. $$\frac{d^n}{dx^n}(uv) = ^nC_0u \frac{d^n}{dx^n} (v)+^nC_1 \frac{d}{dx} (u) \frac{d^{n-1}}{dx^{n-1}} v+...... ^nC_n \frac{d^n}{dx^n} (u)v$$ here $ u=x^2$ and $v=\sin x$ also note $$ \frac{d^n}{dx^n} (\sin x)=sin(x+n\frac{\pi}{2})$$ on solving you get $$f^{n}(x)=x^2\sin(x+n\frac{\pi}{2})+2nx\sin(x+(n-1)\frac{\pi}{2})+n(n-1)\sin (x+(n-2)\frac{\pi}{2})$$

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  • $\begingroup$ wow, very fast downvote (not mine!) $\endgroup$ – Norbert Dec 23 '13 at 12:47
  • $\begingroup$ i dont know why some are interested in downvoting without even noticing the idea behind the answer. $\endgroup$ – Suraj M S Dec 23 '13 at 13:11
  • $\begingroup$ I upvote, and I really don't understand the downvote. $\endgroup$ – Jean-Claude Arbaut Dec 23 '13 at 13:31
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    $\begingroup$ This seems to be a very good answer as it does not use the Taylor's theorem which is probably at a higher level than Leibniz rule for differentiation. I wonder why it is becoming common to down-vote without giving even a few words of comment. $\endgroup$ – Paramanand Singh Dec 23 '13 at 14:00
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The answer you have stated is wrong. It should be $-25*24$. Note that in the Taylor expansion for $f(x)$ the only non-zero term that matters is $-x^{25}/23!$. Differentiate this last expression $25$ terms and plug in zero, you will get

$$f^{(25)}(0) = -25!/23! = -600.$$

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  • $\begingroup$ I think it is $\frac{-x^{25}}{25!}$. $\endgroup$ – lsp Dec 23 '13 at 12:50
  • $\begingroup$ Ok get it ! There's another $x^2$ to be multiplied with ! $\endgroup$ – lsp Dec 23 '13 at 12:52
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Here is an approach. Follow the steps

i) use the identity $\sin x = \frac{e^{ix}-e^{-ix} }{2i} $

ii) use the product rule for differentition.

Note that the function $(x^2)^{(m) }$ vanishes for $m=3$. Chech this related problem.

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