# Fractional Derivatives of Exponential Functions

I am studying the generalization/derivation if the $\alpha^{\mbox{th}}$ derivative of $e^{ax}$. I got lost in the third line. Could someone please please fill in the missing lines so that the derivation will be in detail? I really need to figure this thing out. With $\displaystyle D_x^n f(x)=\lim_{h\to 0}h^{-n}\sum_{m=0}^n(-1)^m{}_nC_m f(x+(n-m)h)$ where $_nC_m=\frac{n!}{m!(n-m)!}$ and with $f(x)=e^{ax}$, \begin{array}{rcl} \displaystyle D_x^\alpha e^{ax}&=& \lim_{h\to 0} h^{-\alpha}\sum_{n=0}^\alpha(-1)^n{} _\alpha C_n e^{a(x+(\alpha-n)h)} \\ &=& e^{ax}\lim_{h\to 0} h^{-\alpha}\sum_{n=0}^\alpha(-1)^n{}_\alpha C_n (e^{ah})^{\alpha-n} \\ &=& e^{ax}\lim_{h\to 0} h^{-\alpha} (e^{ah}-1)^\alpha \mbox{what happened???} \\ &=& a^\alpha e^{ax} \end{array}

I didnt get how the third line came up from the second line. Can you plase fill in the missing details for me? Thank you.

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Binomium of Newton? – Raskolnikov Oct 16 '12 at 8:24
@Raskolnikov Yes, but it is probably better known in the English-speaking world as the binomial theorem. – Harald Hanche-Olsen Oct 16 '12 at 8:39

Try this: $$\sum_{n=0}^{\alpha} \binom{\alpha}{n}(-e^{ah})^{n}=(1-e^{ah})^{\alpha}$$
This is called Binomial theorem. In general: $$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$$