## Calculus: Early Transcendentals (2nd Edition)

$= - \frac{1}{{{{\sin }^2}x}} = - {\csc ^2}x$
$\begin{gathered} show\,\,that\,\,\frac{d}{{dx}}\,\left( {\cot x} \right) = - {\csc ^2}x \hfill \\ \hfill \\ use\,\,the\,\,identity\,\,\,\cot \,x = \frac{{\cos x}}{{\sin x}} \hfill \\ \hfill \\ Use\,\,the\,\,quotiente\,\,rule \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\frac{{\cos x}}{{\sin x}}} \right] = \frac{{\,\left( {\sin x} \right)\,\left( {\cos x} \right)' - \,\left( {\cos x} \right)\,\left( {\sin x} \right)'}}{{{{\sin }^2}x}} \hfill \\ \hfill \\ then \hfill \\ \frac{d}{{dx}}\,\,\left[ {\frac{{\cos x}}{{\sin x}}} \right] = \frac{{\,\left( {\sin x} \right)\,\left( { - \sin x} \right) - \,\left( {\cos x} \right)\,\left( {\cos x} \right)}}{{{{\sin }^2}x}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\cot x} \right] = \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}} \hfill \\ \hfill \\ = - \frac{1}{{{{\sin }^2}x}} = - {\csc ^2}x \hfill \\ \end{gathered}$