I've got a couple of questions regarding derivatives and the arc length formula.
I've been given the arc length formula (where $s$ equals the integral from $x$ to $1$ of $\sqrt{1+(dy/dt)^2}dt$)
I've found an expression for $ds/dx =\sqrt{1+(dy/dx)^2} $, but now I need to show that
$$s = y - x*dy/dx$$
How would I go about that using the original $s$ and $ds/dx$ expressions? I also need to show that $ds/dx$ and the determined $s$ function to show that $x\frac{d^2 y}{dx^2} = \sqrt {1+(dy/dx)^2}$.
For the first part, would I integrate the $s$ integral function?
Your Question is still a little vague, and your sentences are a bit garbled, but I'll answer what I think you're asking. I assume you've been given some form of $y(x)$ that you've not shared here.
So if you use that y(x) to calculate derivs of $y(x)$ you can show that $x\frac{d^2 y}{dx^2} = \sqrt {1+(dy/dx)^2}$, so that part's easy. Then substitute $ds/dx$ for the RHS. If you simply integrate that equation on x, you will get $s = y - x*dy/dx$. Take the deriv wrt x of $s = y - x*dy/dx$ and you'll see what I mean.
If that's not what you needed, you'll have to clarify your Question a lot.