Differentiation of $\sec v$

Let $y = \sec v$. By item 26, §1.1, this may be written $y = \frac{1}{\cos v}$. Differentiating by formula VII,

$$\begin{array}{ll} \frac{dy}{dx} & = -\frac{\frac{d}{dx}(\co... ...} \frac{dv}{dx}\\ & = \sec v \tan v \frac{dv}{dx}. \end{array}$$

Therefore,

$$\frac{d}{dx}(\sec v) = \sec v \tan v \frac{dv}{dx}$$

(equation (XV) in §5.1 above).