Differentiability of $x^\alpha$

Question

How can I show that $x^\alpha$ with $\alpha\in(0,1)$ is differentiable on $(0, +\infty)$?

Usually, I let $x_{0}\in (0,+\infty)$ and compute:

$$\frac{f(x)-f(x_{0})}{x-x_{0}}=\frac{x^\alpha-x_{0}^\alpha}{x-x_{0}}$$

I want the limit of the above term to exist. Any tips on algebraic manipulation in this case?

Answer 1

Let $\alpha=a/b$ be rational. Then $x^\alpha = (x^a)^{1/b}=f(g(x))$. Presumably you've shown that $x^a$ is differentiable. For $f(x)=x^{1/b}$, you can rewrite this as $y=x^{1/b}$ or $y^b=x$. By implicit differentiation this gives $by^{b-1}y'=1$, or in effect $f(x)$ is also differentiable. You can now bound the original limit by arbitrarily close rational approximations of $\alpha$ from above and below.

Answer 2

Another solution: Rewrite $x^{\alpha}$ as $e^{\alpha ln(x)}$ and apply the chain rule.