Answer
$f'(x) = cos (x) - x sin(x)+2sec^{2}(x)$
Work Step by Step
$f(x) = x cos(x) + 2tan (x)$
lets use the formulas $(fg)' = f'g+ fg'$, $(f+g)' = f'+ g'$, $(cf)' = cf'$
$f'(x) = (x cos(x) + 2tan (x))' = (x cos(x))' + (2tan (x))'$
$(2tan (x))'=2(tan (x))' =2sec^{2}(x)$
$(x cos(x))'= x' cos (x) + x (cos(x))' = cos (x) - x sin(x)$
So:
$f'(x) = cos (x) - x sin(x)+2sec^{2}(x)$