I'm in year 11 at high school and have just learn about calculus and the derivative of a function. I found that if I iterated the function $f(x) = x^n$ through the derivative process $k$ number of times where $k \leq n$ and $f(x)$ is $k=1$, $f'(x)$ is $k=2$ I would get: $[n(n-1)(n-2)…(n-(k-2))]x^{n-k}$.
i.e.:
$$
\begin{align}
&k = 1: f(x) = x^n\\
&k = 2: f(x) = nx^{n-1}\\
&k = 3: f(x) = n(n-1)x^{n-2}
\end{align}
$$
The interesting stuff starts to happen at $k=3$, if I expand out the n terms in front of $x$ I get $n^2 – n$. Now If I take the derivative of this I get $2n-1$ and ignoring constants I get $2n$ and if I sub the value for $k$ in (3), I get 6 for both of them as $3^2 – 3 = 9-3 = 6$ and $2*3 = 6$. This is also interesting as this is the factorial for 3. The same thing happens at $k=4$ and onwards, all the derivatives are equal for the value of the $(k-1)$th derivative of $x^n$ and the product of these derivatives are the factorial value for $k$.
My question is: What have I discovered and, if it is interesting or new.
Best Answer
Yes, there is actually a nice general formula (which you should be able to prove by induction):
$$\frac{\mathrm d^k}{\mathrm dx^k}x^n=\frac{n!}{(n-k)!}x^{n-k}=k!\binom{n}{k}x^{n-k}$$