Differentiation of $\csc\, v$

Let $y = \csc\, v$. By item 26, §1.1, this may be written

$$y = \frac{1}{\sin v}.$$

Differentiating by formula VII,

$$\begin{array}{ll} \frac{dy}{dx} & = -\frac{\frac{d}{dx}(\si... ...x}}{\sin^2 v}\\ & = -\csc v \cot v \frac{dv}{dx}. \end{array}$$

Therefore,

$$\frac{d}{dx}(\csc v) = - \csc v \cot v \frac{dv}{dx}$$

(equation (XVI) in §5.1 above).