Differentiation of $\tan v$

Let $y = \tan v$. By item 27, §1.1, this may be written

$$\begin{array}{ll} \frac{dy}{dx} & = \frac{\cos v \frac{d}{d... ...{\frac{dv}{dx}}{\cos^2 v} = \sec^2 v \frac{dv}{dx}. \end{array}$$

Therefore,

$$\frac{d}{dx}(\tan x) = \sec^2 v \frac{dv}{dx},$$

(equation (XIII) in §5.1 above).