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Recall that it is defined by $f(z)=z^c=e^{clog (z)}$, I think we could use the following property

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+ x^{n-3}y^2+ \cdots + y^{n-1})$$

whereby

\begin{align*} \lim_{z \to z_{0}} \frac{f(z)-f(z_{0})}{z-z_{0}}&= \lim_{z \to z_{0}} \frac{z^c-z_{0}^c}{z-z_{0}} \\ &=\lim_{z \to z_{0}} \frac{(z-z_0)(z^{c-1}+z^{c-2}z_0+\cdots + z_{0}^{c-1})}{z-z_{0}} \\ &=\lim_{z \to z_{0}} z^{c-1}+z^{c-2}z_0+\cdots + z_{0}^{c-1} \\ &= z_0^{c-1}+z_0^{c-2}z_0+\cdots + z_{0}^{c-1} \\ &=c z_0^{c-1} \end{align*}

this works when $c$ is a natural value greater than $1$, but what about the remaining cases? is there any way to deal with them? any help would be appreciated!

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1 Answer 1

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If you are allowed to use the chain rule, one gets (at least in the real case): \begin{align*} f(x) := x^{\alpha} = \exp\left(\alpha\ln(x)\right) \Rightarrow f'(x) = \frac{\alpha\exp\left(\alpha\ln(x)\right)}{x} = \alpha x^{\alpha - 1} \end{align*}

Hopefully this helps!

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