Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 19

Show that $\displaystyle \frac{d}{dx} (\cot x) = - \csc^2 x$

Get the reciprocal of $\cot x$

$\displaystyle \cot \frac{\cos x}{\sin x}$

Use Quotient Rule


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