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


\end{aligned}
\end{equation}
$