I was trying to derive this problem for quite some time and I was unable to do it. As a result, I checked the solution on how to do it but there is one step that doesn't make sense.
Question: An object moves so that its velocity, v, is related to its position, s, according to $v=\sqrt{2gs+b^2}$, where b and g are constants. Show that the acceleration of the object is constant.
Solution: $$v=\sqrt{b^2+2gs} \tag{1}$$ $$v=(b^2+2gs)^{1/2} \tag{2}$$ $$\dfrac{dv}{dt}=\dfrac{1}{2}(b^2+2gs)^{-1/2}\;\times\; (0+2g\dfrac{ds}{dt}) \tag{3}$$ $$a=\dfrac{1}{2v}\;\times\; (2gv) \tag{4}$$ $$a=g \tag{5}$$ The part I don't understand is the transition from$\;(3)\;$to $\;(4)\;$. How does the derivative of $\dfrac{ds}{dt} = v$? That line makes no sense.
By definition — the velocity $v$ is the derivative $\frac{ds}{dt}$ of the position function $s$.