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Let $ \ f(x) \ = \ x^4 e^x \ $ . Determine the nth derivative of $ \ f \ $ at $ \ x \ = \ 0 \ $.

I know by working it out that the first, second, and third derivative will be 1. The fourth, fifth, and onward derivatives will be 0.

However, the textbook answer is n(n-1)(n-2)(n-3). I'm having difficulty understanding this, because according to that formula, the first, second, and third derivative will all equal zero, or am I misinterpreting it somehow?

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  • $\begingroup$ I had to withdraw my earlier comment. The first few derivatives equal zero at $ \ x \ = \ 0 \ $ , but this does not continue to be the case. $\endgroup$ – colormegone Jun 2 '15 at 3:32
  • $\begingroup$ Hint: Maclaurin Series. $\endgroup$ – Braindead Jun 2 '15 at 3:32
  • $\begingroup$ Ah, in that case, I'm going to withdraw my previous comment to make myself look like less of an idiot haha $\endgroup$ – Elie Fraser Jun 2 '15 at 3:33
  • $\begingroup$ The power series approach is efficient, but is this a first- or second-semester calculus course? $\endgroup$ – colormegone Jun 2 '15 at 3:35
  • $\begingroup$ Um, I'm not taking a course - I'm studying by working through this study manual ACTEX (P) By Broverman. This is just one question I didn't get in the first problem set. T_T I have a long way to go $\endgroup$ – Elie Fraser Jun 2 '15 at 3:38
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How much calculus do you know? The Taylor series of $e^x$ is $$\sum_{n=0}^\infty \frac{1}{n!} x^n$$ and the Taylor series of $x^4 e^x$ is thus $$\sum_{n=0}^\infty \frac{1}{n!} x^{n+4} = \sum_{n=4}^\infty \frac{1}{(n-4)!}x^n = \sum_{n=4}^\infty \frac{n(n-1)(n-2)(n-3)}{n!} x^n.$$

Since the first four coefficients of the Taylor series are zero, you have $$f(0) = f'(0) = f''(0) = f'''(0) = 0.$$ The remaining derivatives are given by he numerator in the Taylor expansion: $$f^{(n)}(0) = n(n-1)(n-2)(n-3).$$

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Hint :Notice that $f(x)=x^{4}e^{x}=\sum\limits_{n=0}^{\infty}\frac{x^{n+4}}{n!}=\sum\limits_{n=4}^{\infty}\frac{x^{n}}{(n-4)!}$

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  • $\begingroup$ Thank you! This makes sense to me after reading Umberto's post. $\endgroup$ – Elie Fraser Jun 2 '15 at 3:44
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Note that you are only finding the derivative of $f$ at the point $x=0$. You will find that these derivatives have some $x$ term multiplying it for the first three derivatives, so it makes sense that the derivative is $x$ at $0$.

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  • $\begingroup$ Oh, okay! Thank you, this makes sense to me now. $\endgroup$ – Elie Fraser Jun 2 '15 at 3:35

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