Elementary proof that monotone functions are differentiable somewhere

It is well-known that every monotone function $f : \mathbb{R} \to \mathbb{R}$ is differentiable almost everywhere (with respect to Lebesgue measure). It is also known if $E$ has measure $0$, then there exists a continuous, monotone function that is differentiable at no point of $E$.

The proofs of these results, at least those I have seen, are a bit too technical for first-year calculus students to digest. On the other hand, I'm willing to settle for a much weaker result:

Every monotone function is differentiable at some point.

Is there an elementary way avoiding all measure theory, and preferably also avoiding Baire's theorem or other topological concepts that won't be familar to most calculus students showing this?

Edit: to clarify, I'd like to avoid integrals too. See the motivating example further down. If we have the Riemann integral at our disposal, there are much simpler ways to define $x^y$ which gives differentiability with less effort.

If you need to assume continuity to simplify the proof, that's ok.

(One possible) motivation

Let's try to define exponentiation $x^y$ for $x >0$, $y \in \mathbb{R}$.If $y$ is a positive integer, of course $$ x^n = \underbrace{x \cdot x \cdots x}_{n~\text{times}}. $$ Assuming we have dealt with $q$:th roots of real numbers, the extension to rational exponents is also straight-forward: $$ x^{p/q} = \big(\sqrt[q]{x}\big)^p $$ Finally, it's a little tedious, but not too bad to extend first to negative rational numbers $x^{-r} = 1/x^r$ and finally to real exponents by "continuity". Doing all this will give (for a fixed $x$) a continuous monotone function $f(y) = x^y$ satisfying the functional equation $$f(y_1+y_2) = f(y_1)f(y_2).$$ How do we prove that this function $f$ is differentiable? (See Show $\lim\limits_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log for an expanded version of this question.) Among the answers is a clever way to do it using convexity, but I'm still curious if it's possible to give an elementary solution by just exploiting monotonicity.

If we can show that $f$ is differentiable at a single point, then the functional equation implies diffentiability everwhere.

Only a partial answer, for your functional equation $f(y_1+y_2)=f(y_1)f(y_2)$ assuming that $f$ is continuous (and not the zero function): Put $\displaystyle F(x)=\int_0 ^x f(t)dt$. Then $$F(x+y)-F(x)=\int_x^{x+y}f(t)dt=\int_0^y f(t+x)dt=f(x)F(y)$$ Now if you fix a $y$ such that $F(y)$ is not $0$, you have that $f(x)$ is differentiable, as $F$ is differentiable.