Differentiation of ${\rm arcsec}u$

Let $y =\ {\rm arcsec}v$; then $v =\ \sec y$. This function is defined for all values of $v$ except those lying between $-1$ and $+1$, and is seen to be many-valued. To make the function single-valued, $y$ is taken as the arc of smallest numerical value whose secant is $v$. This means that if $v$ is positive, we confine ourselves to points on arc $AB$ (Figure 5.9), $y$ taking on values between 0 and $\frac{\pi}{2}$ (0 may be included); and if $v$ is negative, we confine ourselves to points on arc $DC$, $y$ taking on values between $-\pi$ and $-\frac{\pi}{2}$ ($-\pi$ may be included).

Figure: The inverse secant $\sec^{-1}\ x$ using SAGE.

Figure 5.10: The standard branch of ${\rm arcsec}\x$ using SAGE.

Differentiating with respect to $y$ by IV, $\frac{dv}{dy} =\ \sec y \tan y$; therefore $\frac{dy}{dv} = \frac{1}{\sec y \tan y}$, 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

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

(since $\sec y = v$, and $\tan y = \sqrt{\sec y - 1} = \sqrt{v^2 - 1}$, the plus sign of the radical being taken, since $\tan y$ is positive for an values of $y$ between 0 and $\frac{\pi}{2}$ and between $-\pi$ and $-\frac{\pi}{2}$, including 0 and $-\pi$). Therefore,

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

(equation (XXII) in §5.1 above).