Definition of Curvature

Question

Derivation of Definition of Curvature

$$ \phi = tan^{-1}(y')$$

1. Differentiate with respect to s.

$$ \frac {d \phi} {ds} = \frac d {ds} (tan^{-1} y')$$

2. Chain rule.

$$ \frac {d \phi} {ds} = \frac d {dx} (tan^{-1}y') \frac {dx} {ds}$$

3. Chain rule.

$$ \frac {d \phi} {ds} = \frac d {d^2y} (tan^{-1}y') \frac {d^2y} {dx^2} \frac {dx} {ds}$$

The remaining (unlisted) steps I understand, but I have trouble applying the chain rule the second time.

Here is my attempt to perform the chain rule the second time:

$$ \frac {d \phi} {ds} = \frac d {dx} (tan^{-1}y') \frac {dx} {ds} \frac {d^2y} {d^2y}$$

$$ \frac {d \phi} {ds} = \frac d {d^2y} (tan^{-1}y') \frac {dx} {ds} \frac {d^2y} {dx}$$

Where does the additional $\frac 1 {dx}$ come from?

Answer 1

Let $g=\frac{dy}{dx}$ so that $\phi = \tan^{-1}(g)$. We make this substitution to relieve us from the abundance of differentials! Then

$$\frac {d \phi} {ds} = \frac {d}{dx}\left[\tan^{-1}(g)\right]\frac{dx}{ds}$$ by the chain rule. But by the chain rule again, $$\frac {d}{dx}\left[\tan^{-1}(g)\right]=\frac{d}{dg}\left[\tan^{-1}(g)\right]\frac{dg}{dx}.$$ Putting those together, $$\frac {d \phi} {ds} = \frac{d}{dg}\left[\tan^{-1}(g)\right]\frac{dg}{dx}\frac{dx}{ds}.$$ But remember our substitution, $g=\frac{dy}{dx}=y'$. So, $$\frac {d \phi} {ds} = \frac{d}{d(y')}\left[\tan^{-1}(y')\right]\frac{d}{dx}\left[\frac{dy}{dx}\right]\frac{dx}{ds},$$ which can be written as $$\frac {d \phi} {ds} = \frac{d}{d^2y}\left[\tan^{-1}(y')\right]\frac{d^2y}{dx^2}\frac{dx}{ds}.$$