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}
$