Why is there this recurring pattern in the derivative of $x\cdot e^{-k\cdot x}$?

Question

I was doing an exercise for which I needed to take the derivative(s) of the function

$$f_k(x)=x\cdot e^{-k\cdot x}$$

Doing this a few times, I noticed some sort of recurring pattern. I concluded that one can write the $n$-th derivative of the function as

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}((-1)^nkx-(-1)^n\cdot n)$$

or perhaps, to improve readability

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(-kx+n)$$ for odd $n$ and

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(kx-n)$$ for even $n$.

While I know how to compute each derivative, I do not understand why the derivative follows this pattern-like structure. Is there any intuitive idea behind this?

Clarification: By "intuitive", I mean anything that does not purely rely on blindly computing each integral. A very beautiful way to represent derivatives visually is shown in this 3blue1brown video; I don't know if anything like this is possible to explain the above described pattern. Basically, I am looking for something which actually explains why there is a pattern instead of just noticing it.

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_{Note: This is not the actual homework assignment, it just came up when doing it.}

Answer 1

Introduce an auxiliary variable $\lambda$, and put $$f(x,\lambda):=xe^{\lambda x}={\partial\over\partial\lambda}\bigl(e^{\lambda x}\bigr)\ .$$ The exchangeability of partial derivatives then implies $${\partial^n\over\partial x^n}f(x,\lambda)={\partial\over\partial\lambda}{\partial^n\over\partial x^n}\bigl(e^{\lambda x}\bigr)={\partial\over\partial\lambda}\bigl(\lambda^n e^{\lambda x}\bigr)=\lambda^{n-1}(n+\lambda x)e^{\lambda x}\ .$$ When $\lambda:=-k$ we have your formulas.

Answer 2

The reason behind the pattern relies merely on the taylor expansion of the exponential: $$e^{x}=\sum_{n=0}^{+\infty}\frac{x^{n}}{n!}$$ and if you take it with $-k$ in the exponent then: $$e^{-kx}=\sum_{n=0}^{+\infty}(-1)^{n}k^{n}\frac{x^{n}}{n!}\implies xe^{-kx}=\sum_{n=0}^{+\infty}(-1)^{n}k^{n}\frac{x^{n+1}}{n!}$$ Now, if you consider the $n-th$ derivative of $xe^{-kx}$ you will see it is exactly the $n-th$ derivative of the series here above. As for intution, it is actually a weird question. Keep in mind that, for instance, in nature the concept of "continous function" doesn't really make sense: all the measures you will take in your life will be discrete quantities. The idea of continous quantity it's somthing that helps us when doing calculus. The interpretetion given by 3Blue1Brown is indeed appealing but it is actually unprecise: $dx$ is not an "infinitesimal" quantity, not since we moved from the interpretetion Leibniz gave of calculus when he basically invented it. You should see in $dx$ a way of $\textit{measure}$ something. In mathematics we measure sets, and if you will deepen your knowledge about measure theory you will see what it means exactly. Moreover you will see in which sense $dx$ is just a measure.

Answer 3

Following the hint of @HansLundmark is quite convenient.

The general Leibniz rule gives us \begin{align*} \color{blue}{\frac{\partial^n}{\partial x^n}}\color{blue}{\left(x e^{-k\cdot x}\right)} &=\sum_{j=0}^n\binom{n}{j}\left(\frac{\partial^j}{\partial x^j} x\right)\left(\frac{\partial^{n-j}}{\partial x^{n-j}} e^{-k x}\right)\tag{1}\\ &=\binom{n}{0}x\left(\frac{\partial^{n}}{\partial x^{n}}e^{-k x}\right)+\binom{n}{1}\left(\frac{\partial^{n-1}}{\partial x^{n-1}}e^{-k x}\right)\tag{2}\\ &=x\left(-k\right)^ne^{-k x}+n\left(-k\right)^{n-1}e^{-k x}\\ &\,\,\color{blue}{=(-1)^nk^{n-1}e^{-kx}\left(kx-n\right)} \end{align*} Here we do not have to look for a common pattern, since the Leibniz rule gives us the wanted expression right at hand.

Comment:

• In (1) we get by the general Leibniz rule more or less the final expression without the need to calculate derivatives iteratively.

• In (2) we use the nice fact that $\left(\frac{\partial^j}{\partial x^j} x\right)=0$ if $j>1$ so that only two summands give a contribution.