Are you math genius? Let us see if you can answer this one?

f ' (x) = (d(secx + tanx)^2 / d(secx + tanx) ) / ( (dsecx/dx) + (dtanx/dx)) (by "chain rule" and " sum rule")

f ' (x) = 2(secx + tanx) ( ((secx)^2) + (sectanx)) ( using standard first order derivatives of secx and tanx and remembering that d(y^2) /dy = 2y )

f ' (x) =2 (secx + tanx) (secx + tanx) secx = 2 f(x) secx

f " (x) = 2 f ' (x) secx + 2 f(x) ( dsecx / dx) (by the "product rule")
=2 f ' (x) secx + 2 f(x) secx tanx
= 2 f ' (x) secx + f ' (x) tanx = f ' (x) (2secx + tanx) ( since as proved previously 2f(x) secx =f'(x) )

REMARKS :- THE PROBLEMS WERE EASY AND TRIVIAL.