Answer
\[\frac{{dy}}{{dx}} = {\sec ^2}x - {\csc ^2}x\]
Work Step by Step
\[\begin{gathered}
y = \tan x + \cot x \hfill \\
\hfill \\
differentiate \hfill \\
\hfill \\
\frac{d}{{dx}} = \frac{d}{{dx}}\,\,\left[ {\tan x} \right] + \frac{d}{{dx}}\,\,\left[ {\cot x} \right] \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {\sec ^2}x - {\csc ^2}x \hfill \\
\end{gathered} \]