Question

y=5sinx+tgx-2
Y'= ??????????

Answer 1

$y=5sinx+tgx-2 \ \ y'=(5sinx+tgx-2)'=(5sinx)'+(tgx)'-(2)'= \ \ = 5cosx+frac{1}{cos^2x}-0=5cosx+frac{1}{cos^2x}$