Let f(x)=5secx+4cosx
Thus using first principle,
f′(x)=limh→0f(x+h)−f(x)h
= limh→05sec(x+h)+4cos(x+h)−[5secx+4cosx]h
= 5limh→0[sec(x+h)−secx]h+4limh→0[cos(x+h)−cosx]h
= 5limh→01h[1cos(x+h)−1cosx]+4limh→01h[cos(x+h)−cosx]
= 5limh→0[cosx−cos(x+h)cosxcos(x+h)]+4limh→01h[cosxcosh−sinxsinh−cosx]
= 5cosx⋅limh→01h⎡⎢
⎢⎣−2sin(2x+h2)sin(−h2)cos(x+h)⎤⎥
⎥⎦+4[−cosxlimh→0(1−cosh)h−sinxlimh→0sinhh]
= 5cosx⋅limh→0⎡⎢
⎢
⎢
⎢
⎢
⎢⎣sin(2x+h2)⋅sin(h2)h2cos(x+h)⎤⎥
⎥
⎥
⎥
⎥
⎥⎦+4[(−cosx)⋅(0)−(sinx)⋅1]
= 5cosx⋅⎡⎢
⎢⎣limh→0sin(2x+h2)cos(x+h)⋅limh→0sin(h2)h2⎤⎥
⎥⎦−4sinx
= 5cosx⋅sinxcosx⋅1−4sinx
= 5secxtanx⋅−4sinx