I have this equation, $f(x) = e^{-x^2}$. My question is how should I find $f^{(10)} (0)$, ie the tenth derivative of this equation.

I have tried differentiating to get a formula, and I get $f^{(n)}(0) = (-1)^n*x^n*e^{-x^2}$. However, substituting $n=10$ and $x=0$ into this general formula gives me an answer of $0$. But this is incorrect as the answer given is $-10!/5!$.

Should I convert this equation into a Taylor Series? If yes, how should I use Taylor Series to find the derivative?



We know that: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!}+ \frac{x^4}{4!} + \frac{x^5}{5!} + o(x^5)$$ Using $-x^2$ instead of $x$, we obtain: $$e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!}+ \frac{x^8}{4!} - \frac{x^{10}}{5!} + o(x^{10})$$

The general term on the Taylor Series is $\dfrac{f^{(k)}(0)}{k!}x^k$. The term with $x^{10}$ has a coefficient of $\dfrac{-1}{5!}$, hence: $$\frac{f^{(10)}(0)}{10!} = -\frac{1}{5!} \implies f^{(10)}(0) = -\frac{10!}{5!}$$ as desired.

  • $\begingroup$ How would you express it if you were substitute -x^2 for x in the standard summation formula for e^x? (sorry I don't know how to do the formatting for summation so I'm not going to type it out here) $\endgroup$ – thbcm Aug 3 '14 at 18:56
  • $\begingroup$ I assume that you're referring to $e^x = \sum_{k = 0}^{+\infty} \frac{x^k}{k!}$ The first expression for $e^x$ I wrote is exactly the same as this summation. The only difference is that I didn't use in face the series, just the expansion until $5$th order (which would become the $10$th order we needed for $e^{-x^2}$). Going further than this wasn't needed to solve this specific problem. $\endgroup$ – Ivo Terek Aug 3 '14 at 19:02
  • $\begingroup$ Ah I get it now, thanks! :) $\endgroup$ – thbcm Aug 3 '14 at 19:18
  • $\begingroup$ @inggumnator protip: if the answer did help, it is nice to accept it (see the checkmark below the up/down votes on the answer). A lot of people here do not answer questions from users that do not give an appropiate feedback. It might be useful to you in the future. $\endgroup$ – Ivo Terek Aug 17 '14 at 17:30
  • $\begingroup$ Right, I've done that! Sorry about that! $\endgroup$ – thbcm Aug 25 '14 at 3:58

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