I take the definition of arc length of a smooth curve between $x=a$ and $x=b$ to be: $$\int^a_b \sqrt{1+f'(x)^2}dx$$ Then how do I derive the formula for polar coordinates?
And please don't use infinitesimals, (or poorly defined things). Only use rigorous arguments like epsilons and deltas.
If you parameterize the curve using $x=x(t)$, $y = y(t)$, $a \le t \le b$ the formula for arclength is $$L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2} \, dt.$$
You can let $r(t)$, $\theta(t)$ denote the polar coordinates of the point $(x(t),y(t))$.
Since $x(t) = r(t) \cos \theta(t)$ and $y(t) = r(t) \sin \theta(t)$ you get $$x'(t) = r'(t) \cos \theta(t) - r(t) \sin \theta(t) \theta'(t)$$and $$y'(t) = r'(t) \sin \theta(t) + r(t) \cos \theta(t) \theta'(t).$$ Square to get $$x'(t)^2 = r'(t)^2 \cos^2 \theta(t) - 2 r(t) r'(t) \cos \theta (t) \sin \theta(t) \theta'(t) + r(t)^2 \sin^2 \theta(t) \theta'(t)^2$$ $$y'(t)^2 = r'(t)^2 \sin^2 \theta(t) + 2 r(t) r'(t) \cos \theta (t) \sin \theta(t) \theta'(t) + r(t)^2 \cos^2 \theta(t) \theta'(t)^2$$ thus $$x'(t)^2 + y'(t)^2 = r'(t)^2 + r(t)^2 \theta'(t)^2$$ so that $$L = \int_a^b \sqrt{r'(t)^2 + r(t)^2 \theta'(t)^2} \, dt.$$