# $n \text{th}$ derivative of $f(x)$?

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?

• 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. – colormegone Jun 2 '15 at 3:32
• Hint: Maclaurin Series. – Braindead Jun 2 '15 at 3:32
• Ah, in that case, I'm going to withdraw my previous comment to make myself look like less of an idiot haha – Elie Fraser Jun 2 '15 at 3:33
• The power series approach is efficient, but is this a first- or second-semester calculus course? – colormegone Jun 2 '15 at 3:35
• 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 – Elie Fraser Jun 2 '15 at 3:38

## 3 Answers

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: $$f^{(n)}(0) = n(n-1)(n-2)(n-3).$$

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

• Thank you! This makes sense to me after reading Umberto's post. – Elie Fraser Jun 2 '15 at 3:44

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 $0$.

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