The General Leibniz Rule is given as
$$(uv)^{(n)}=\sum_{k=0}^n\binom nku^{(n-k)}v^{(k)}$$
Where $u^{(n)}$ means the $n$th derivative of $u$ with respect to $x$ and $\binom nk$ is a binomial coefficient.
I tried to generalize this for fractional calculus, making a few assumptions. Firstly, I imagined it'd turn out like Euler's generalized binomial expansion:
$$(uv)^{(n)}=\sum_{k=0}^\infty\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}u^{(n-k)}v^{(k)}$$
Nicely, if we define this for $n=-1$, we get a series that should be equal to integration by parts.
$$(uv)^{(-1)}=\sum_{k=0}^\infty(-1)^ku^{(-k-1)}v^{(k)}$$
Assuming of course that $u^{(-k-1)}v^{(k)}$ tends to $0$.
Also, if we allow $u=1$, we get a formula that allows calculations of fractional derivatives using integer ordered derivatives.
$$1^{(n)}=\frac{x^{-n}}{\Gamma(1-n)}$$
$$f^{(n)}(x)=\sum_{k=0}^\infty\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)\Gamma(k-n+1)}x^{k-n}f^{(k)}(x)$$
$$=\sum_{k=0}^\infty\frac{\Gamma(n+1)}{-(n-k)^2\Gamma(k+1)\Gamma(n-k)\Gamma(k-n)}x^{k-n}f^{(k)}(x)$$
Applying a reflection formula,
$$f^{(n)}(x)=\sum_{k=0}^\infty\frac{\Gamma(n+1)\sin[\pi(n-k)]}{-\pi(n-k)\Gamma(k+1)}x^{k-n}f^{(k)}(x)$$
$$f^{(n)}(x)=\frac{\Gamma(n+1)}{-\pi}\sum_{k=0}^\infty\frac{\sin[\pi(k-n)]}{(k-n)\Gamma(k+1)}x^{k-n}f^{(k)}(x)$$
Through graphing, I noticed that this converges mostly for $x>0$, and possibly converges or diverges everywhere else (my calculator doesn't take an upper limit of infinity.) I took the following well known fractional derivatives to test my formula, mainly $x^a$, $a^x$, and $\sin(x)$. For all of these, convergence for $x<0$ was somewhat bad, and for $x>0$, it was good. I note that sometimes, an upper limit of $10$ in the summation was sufficient for decently small $x$.
And of course, this is undefined for $n=k$. If $n=k$, correction is possible by having $\lim_{x\to0}\frac{\sin(x)}x=1$.
Question I wonder if my formula could be further reduced and if it has already been discovered.
6/15/16 I do think my graphing calculator lacks the ability to graph for $x<0$ due to $x^{k-n}$ where $k-n$ being a fractional number... would be nice if someone could make a complex graph of the thing. If you unsure of what to graph, try inputting $f^{(k)}(x)=(\ln(a))^ka^x$ and comparing the graph of one side to the other.