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Of course, I'm not really familiar with all fractional derivative methods, but is it a necessary rule that they all should comply with? If not, which ones, for example, do and which don't ?

( Specifically: I am interested in Caputo's fractional derivation method $$D^\alpha_+ f(x)=1/\Gamma(n-\alpha)\int^x_0(x-\zeta)^{n-\alpha-1}f^{(n)}(\zeta)d\zeta$$ where $n=[\alpha]+1$.

More info. : In Caputo's method, taking the half-derivatve of an exponential function is : $$D^{1/2}_+ \exp(kx)=k^{1/2}\exp(kx)*(1-\Gamma(1/2,kx)/\Gamma(1/2)).$$
Now, say I wish to take another derivative of the above, meaning $$D^{1/2}_+ [D^{1/2}_+ \exp(kx)]=?.$$ Can I use the $[fg]'=f'g+g'f$ rule for that? Can I simply say $$D^{1/2}_+ D^{1/2}_+f(x)=f'(x) \implies D^{1/2}_+ D^{1/2}_+\exp(kx)=k*\exp(kx) ?$$ Or am I really in truble ?)

Siminore
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user1611107
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    As far as I know, the most terrible issue of fractional derivatives is precisely that they do not obey the Leibnitz rule. – Siminore Jan 28 '14 at 12:45
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    This paper may be a useful reference to you. See Remark 1 and the surrounding text.

    http://www1.beuth-hochschule.de/~luchko/papers/FreedFordDiethelmLuchko.pdf

     – JessicaK Jan 28 '14 at 12:53
  • @JessicaK: Thanks for the paper. I'll go over it. – user1611107 Jan 28 '14 at 13:16
  • @JessicaK: See my last edit to the question, do I really need such a 'modified Leibniz rule' as the one that they suggest in the paper, for the case of consecutive derivation that I wrote above? Or is there a simpler way to calculate higher order fractional derivatives? – user1611107 Jan 28 '14 at 13:27

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For the second rule, no. For the first rule, yes, but it becomes infinite series rather than finite expression.

Anixx
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