Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 17

Show that $\displaystyle \frac{d}{dx} (\csc x) = - \csc x \cot x$

Get the reciprocal of $\csc x$

$\csc x = \displaystyle \frac{1}{\sin x}$

Use the Quotient Rule to derive $\displaystyle \frac{1}{\sin x}$


$\begin{equation} \begin{aligned} \frac{d}{dx} (\csc x) =& \frac{\displaystyle \sin x \frac{d}{dx} (1) - \left[ 1 \frac{d}{dx} (\sin x) \right]}{(\sin x)^2} && \text{} \\ \\ \frac{d}{dx} (\csc x) =& \frac{\sin x (0) - (\cos x)}{\sin ^2 x } && \text{Simplify the equation} \\ \\ \frac{d}{dx} (\csc x) =& \frac{- \cos x}{\sin ^2 x} && \text{Factor out $\displaystyle \frac{-1}{\sin x}$ and $\displaystyle \frac{\cos x}{\sin x}$} \\ \\ \frac{d}{dx} (\csc x) =& - \left( \frac{1}{\sin x} \right) \left( \frac{\cos x}{\sin x}\right) && \text{Get their identities} \\ \\ \frac{d}{dx} (\csc x) =& - \csc x \cot x && \text{} \end{aligned} \end{equation}$