how do i differentiate √secx+tanx ??

Answer 1

I could never remember those formulas, so

f= (cosx)^-1/2 + (sinx)(cosx)^-1

f'= .5 sinx*(cosx)^-3/2 + cosx*(cosx)^(-1)+1*sinx(sinx)cosx)^-2

and that can be simplified. check my work.

Answer 2

or

y = (secx)^1/2 + tanx

dy/dx = (1/2)(secx)^(-1/2)((secx)tanx) + sec^2 x
= (-1/4)(tanx)(secx)^1/2 + sec^2 x