Differentiation of ${\rm arcvers}\, v$

Let^5.12 $y = {\rm arcvers}\, v$; then $v = {\rm vers}\, y$. Differentiating with respect to $y$ by XVII,

$$\frac{dv}{dy} =\ \sin y;$$

therefore $\frac{dy}{dv} = \frac{1}{\sin y}$, by (5.2). But since $v$ is a function of $x$, this may be substituted in the formula $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$, by (5.1).giving

$$\frac{dy}{dx} = \frac{1}{\sin y} \cdot \frac{dv}{dx} = \frac{1}{\sqrt{2v - v^2}} \frac{dv}{dx}$$

(since $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - (1 - {\rm vers}\, y)^2} = \sqrt{2v - v^2}$, the plus sign of the radical being taken, since $\sin\, y$ is positive for all values of $y$ between 0 and $\pi$ inclusive). Therefore,

$$\frac{d}{dx} (\operatorname{arcvers}\, v) = \frac{\frac{dv}{dx}}{\sqrt{2v - v^2}}$$

(equation (XXIV) in §5.1 above).