I'm looking for a non-piecewise function -- $g(n,x)$ -- that satisfies this equation:
$g(n,x)=\large\frac{d^{n}}{dx^{n}}x^{r}$
Where $n\in\Bbb{Z}$ and is the $n^{th}$ derivitive of $f(x)$
I think I've found $g(n,x)$, but I'm not sure. This is what I've come up with
$\large\frac{d^{n}}{dx^{n}}x^{r} =\large\large\frac{x^{r-n}r!}{(r-n)!}$
Below is my reasoning
The power rule for derivatives, states that
$\frac{d^{}}{dx^{}}x^{r}=r(x^{r-1})$
A natural consequence is that
$\frac{d^{2}}{dx^{2}}x^{r}=r(r-1)(x^{r-2})$
and...
$\frac{d^{3}}{dx^{3}}x^{r}=r(r-1)(r-2)(x^{r-3})$
Following this pattern gets you my original equation
Can anyone verify weather or not this is correct or not? If so, would this work for all values of $r$ and $n$, including when $n < 0$ (the $n^{th}$ integral)? I don't know if there is a way to use negative factorials, but if there is, would this still hold true?
Also, I'm wondering if there a more generalized solution to this. Any help would be appreciated.
$$ \frac{{\rm d}^n}{{\rm d}x^n} x^r = \frac{r!}{(r-n)!} x^{r-n} $$
You are correct
$$ = r (r-1) (r-2) \ldots (r-n+1) \,x^{r-n} $$
Run a few test cases through Wolfram Alpha and you can confirm this yourself.