Answer
$\displaystyle y' = 4tanx\ secx - cosec^{2}x$
Work Step by Step
$\displaystyle y = \frac{4}{cosx} + \frac{1}{tanx}$
$\displaystyle y' = \frac{cosx(0) + 4sinx}{cos^{2}x} + \frac{tanx(0) - sec^{2}x}{tan^{2}x}$
$\displaystyle y' = \frac{ 4sinx}{cos^{2}x} + \frac{- sec^{2}x}{tan^{2}x}$
$\displaystyle y' = 4tanx\ secx - cosec^{2}x$
