If $s(x)= \int \sqrt{1+ \Big(\frac{dy}{dx}\Big)^2} dx$, what is $x(s)$? $s(x)$ is the formula for arc length of a function $f(x)=y$. 
In the book I'm studying, curvature is defined as the instantaneous rate of change of direction (inclination of angle $\theta$) with respect to arc length, where $\theta = \arctan\big(\frac{dy}{dx}\big) $, all of which makes sense to me.
In order to find the curvature function, or $\frac{d\theta}{ds}$, the author goes:
$$\frac{d\theta}{ds} = \frac{d}{ds} \left(\arctan\left(\frac{dy}{dx}\right)\right)$$
Followed by: Since $\theta$ is a function of $x$, to get $\frac{d\theta}{ds}$ we have to use the chain rule:
$$\frac{d\theta}{ds} = \frac{d}{dx}\left(\arctan \left(\frac{dy}{dx}\right)\right) \cdot \frac{dx}{ds}$$
where $$\frac{dx}{ds} = \frac{1}{\frac{ds}{dx}} = \frac{1}{\sqrt{1+ \big(\frac{dy}{dx}\big)^2}}$$
All of this makes sense, even intuitively to some extent. However, I'm unsatisfied by the "skipping" of the original function.
In other words, I'd like to have an expression for $\theta(s)$, from which I could directly take the derivative $\frac{d\theta}{ds}$.
But I can't figure out how to do it and I get lost between all of the different functions and how they interact, and I think my main problem is that I have no idea how to get an expression for $x(s)$
 A: In differential geoemetry, most of the calculation needs you parametrize the curve by arclength. This means you have to reparametrize your curve such that it has unit speed. 
In this case, your curve can be thought as: $\gamma(x):[a,b]\rightarrow\mathbb{R}^2:x\mapsto(x,f(x))$. It doesn't have the unit speed, so we have to reparametrize the curve by a new parameter say $s$ with $\gamma(x(s)):[c,d]\mapsto\mathbb{R}^2$ such that $|\gamma'(x)x'(s)|=1$. The   parameter $s$ may not unique, but it has to satisfy the condition $|\gamma'(x)x'(s)|=1$. This is why your book let $s(x)=\int_a^x\sqrt{1+(f'(x))^2}dx$ since it's easy to check this condition is satisfied.
So if you really want to write $\theta$ as a function of $s$. It should be something like this $\theta(x(s))=\arctan f'(x(s))$. You know $x$ is a function of $s$ and $|x'(s)|=1/|\gamma'(x)|=1/\sqrt{1+(f'(x))^2}$.
A: If you simplify the expression for $\frac{d\theta}{ds}$,
you will see $$\kappa=\frac{f''}{(1+(f')^2)^{3/2}}$$
If you want to find the expression explicitly for $\phi(s)=\theta(x(s))$, you see $\theta(x)=arctan(\frac{dy}{dx})$, you will need to find $x(s)$, which is a function that takes in arclength and output the value of x.
To find it, you will need arclength parametrization. Assume the function starts at $x=a$ and be defined for $x\geq a$, let  $\bar{r}(x)$ be a vector function and $\bar{r}(x)=(x,f(x))$
And
$$s(x)=\int_a^x|\bar{r}'(t)|dt=\int_a^x \sqrt{1+f'^2}dt$$ is a function of arclength
Let $F(x)$ be the antiderivative of $$\int \sqrt{1+f'^2}dx$$ (which may or may not exist), 
Then $F(x)-F(a)=s(x)$,
$$s(x)+F(a)=F(x)$$
$$F^{-1}(s+F(a))=x(s)$$ since F(x) must be bijective (but may not have a close form)
So by using this construction you can define $x(s)$ (know as arc length paramrtrization/ natural parametrization) to find $\theta(x(s))=\phi(s)=arctan(f'(x(s))$ 
