Let f(x)=2tanx−7secx
Thus using first principle,
f′=limh→0f(x+h)−f(x)h
= limh→01h[2tan(x+h)−7sec(x+h)−2tanx+7secx]
= limh→01h[2{tan(x+h)−tanx}−7{sec(x+h)−secx}]
= 2limh→01h[tan(x+h)−tanx]−7limh→01h[sec(x+h)−secx]
= 2limh→01h[sin(x+h)cos(x+h)−sinxcosx]−7limh→01h[1cos(x+h)−1cosx]
= 2limh→01h[sin(x+h)cosx−sinxcos(x+h)cosxcos(x+h)]−7limh→01h[cosx−cos(x+h)cosxcos(x+h)]
= 2limh→01h[sin(x+h−x)cosxcos(x+h)]−7limh→01h⎡⎢
⎢⎣−2(x+x+h2)sin(x−x−h2)cosxcos(x+h)⎤⎥
⎥⎦
= 2limh→0[(sinhh)1cosxcos(x+h)]−7limh→01h⎡⎢
⎢⎣−2sin(2x+h2)sin(−h2)cosxcos(x+h)⎤⎥
⎥⎦
= 2(limh→0sinhh)(limh→01cosxcos(x+h))−7⎛⎝limh2→0sinh2h2⎞⎠⎛⎜
⎜⎝limh→0sin(2x+h2)cosxcos(x+h)⎞⎟
⎟⎠
= 2.1. 1cosxcosx−7⋅1(sinxcosxcosx)
= 2sec2x−7secxtanx