Rigorously show that $x\mapsto x^{\alpha}$ is monotonically increasing for all $\alpha>0$. How would one rigorously prove that $x^\alpha$ is monotonically increasing for all $\alpha>0$? I know it should be simple but I'm finding it surprisingly difficult. I tried to argue that $x^\alpha=e^{\alpha \ln(x)}$ and since
$$\frac{\mathrm{d}}{\mathrm{d}x}e^{\alpha \ln(x)}=\frac{\alpha e^{\alpha \ln (x)}}{x}
$$
and the RHS is positive for all $x > 0$, we have that $x^\alpha$ is monotonically increasing. My two questions are: (1) is this argument correct, and (2) is there a more direct proof that does not require rewriting $x^\alpha$ using $e$?
 A: For $x>0$ we have by definition for all $\alpha \in \mathbb{R}$,
$$ x^{\alpha} = \mathrm{e}^{\ln(x)\alpha}$$
Firstly for all $\alpha>0$,
$$x \in \mathbb{R}_+^*\mapsto \alpha\ln(x) $$ is monotonically increasing on $\mathbb{R}_+^*$ because $\ln$ is increasing on $\mathbb{R}_+^*$.
Furthermore, $\exp$ is monotonically increasing on $\mathbb{R}$ therefore by composition, we have
$$x \in \mathbb{R}_+^* \mapsto \mathrm{e}^{\ln(x)\alpha}= x^{\alpha} $$
monotonically increasing on $\mathbb{R}_+^*$.
A: Your reasoning is correct. Note that $x^{\alpha}$ for any $\alpha$ real number is actually defined in this way, so if you start with this definition this is one of the easiest and most direct way to prove it.
A: You are using the geometric meaning of the derivative and it is correct. Another way of solving it is... see what happen(COMPARE) to the image of $x$ and $y$ through $f(z)=z^{\alpha}$ when $0<x<y$. We wished that $x^{\alpha}<y^{\alpha} \iff (\frac{x}{y})^{\alpha}<1$ and this is true once $x<y\iff \frac{x}{y}<1$.
