Differentiation of $\arcsin v$

Let $y = \arcsin\ v$, then $v = \sin\ y$.

It should be remembered that this function is defined only for values of $v$ between $-1$ and $+1$ inclusive and that $y$ (the function) is many-valued, there being infinitely many arcs whose sines will equal $v$. Thus, Figure 5.4

Figure: The inverse sine $\sin^{-1}\ x$ using SAGE.

Figure: A single branch of the function $f(x)=\arcsin(x)$.

represents only a piece of the multi-valued inverse function of $\sin(x)$, represented by taking the graph of $\sin(x)$ and flipping it about the $45^o$ line. In the above discussion, in order to make the function single-valued; only values of $y$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ inclusive are considered; that is, the arc of smallest numerical value whose sine is $v$.

Differentiating $v$ with respect to $y$ by XI, $\frac{dv}{dy} = \cos\ y$; therefore $\frac{dy}{dv} = \frac{1}{\cos y}$, by (5.2). But since $v$ is a function of $x$, this may be substituted in $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$ (see (5.1)),giving

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

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

$$\frac{d}{dx}(\arcsin\, v) = \frac{\frac{dv}{dx}}{\sqrt{1 - v^2}}$$

(equation (XVIII) in §5.1 above).