Main Question
Consider some curve $y(x)$, going from a point $(x_0,y_0)$ to a point $(x_1,y_1)$.
Let $L$ be the length of the curve, and the function $F$ be the rate of change of the length of this curve with respect to $x$.
$$\frac{d L}{d x}=F=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$$
This is well demonstrated in the image below:
The arc-length of the entire curve would therefor be given by:
$$\int_{x_0}^{x_1}{F}dx$$
At each $(x,y)$ along the curve, the curve has a slope of $y'(x)$. Consider the derivative: $$\frac{\partial F}{\partial y^{\prime}(x)}$$
This derivative tells us how the rate of growth of the length of our curve with respect to $x$ changes with respect to slope of our function at any point. How $F=\frac{dL}{dx}$ changes with respect to $y'(x)=\frac{dy}{dx}$.
If we evaluate the derivative, we get:
$$\frac{d F}{d y^{\prime}(x)}=\frac{d}{d\left(y^{\prime}(x)\right)}(\sqrt{\left(1+(d y / d x)^{2}\right)})=\frac{(d y / d x)}{\sqrt{1+(d y / d x)^{2}}}=\frac{d y / d x}{dL / dx}=\frac{d y}{d L}$$
This is saying that the rate of change of the rate of change of the path-length with respect to $x$ all with respect to the slope of our function is equal to the rate of change of $y$ with respect to the arc-length.
Why?
This MUST have an intuitive explanation! But, although I've tried and tried, I haven't been able to understand it.
Additional Notes:
If in the the picture above, let the bottom-most left angle of the infinitesimal triangle be $\theta$. Then, the expression for $\frac{d F}{d y^{\prime}(x)}$ can be rewritten as:
$$\frac{d}{d\left(y^{\prime}(x)\right)}(\sqrt{\left(1+(\tan(\theta))^{2}\right)})=\frac{(\tan(\theta))}{\sqrt{1+(\tan(\theta))^{2}}}=\frac{\tan(\theta)}{\sec(\theta)}=\sin(\theta)$$
Sure enough, $\sin(\theta)$ is the rate of change of $y$ with respect to the arc-length. However, this didn't help me understand the result.
Additionally, I would think that if I wanted to get the rate of change of $F$ with respect to $y'(x)$, I could simply divide one by the other, which would give me...
$$\frac{d}{dy^{\prime}(x)}\left(\frac{dL}{dx}\right)=\frac{d L}{dx} \div \frac{dy}{dx}=\frac{dL}{dy}.$$
...the reciprocal of what regular'ol differentiation does. Why?
I appreciate all and any help. Thank you so much.