Find the derivative ,BY FIRST PRINCIPAL!
5secx + 4cosx
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Let f(x)=5secx+4cosx
Thus using first principle,
f
′
(x)=
h→0
lim
h
f(x+h)−f(x)
=
h→0
lim
h
5sec(x+h)+4cos(x+h)−[5secx+4cosx]
= 5
h→0
lim
h
[sec(x+h)−secx]
+4
h→0
lim
h
[cos(x+h)−cosx]
= 5
h→0
lim
h
1
[
cos(x+h)
1
−
cosx
1
]+4
h→0
lim
h
1
[cos(x+h)−cosx]
= 5
h→0
lim
[
cosxcos(x+h)
cosx−cos(x+h)
]+4
h→0
lim
h
1
[cosxcosh−sinxsinh−cosx]
=
cosx
5
⋅
h→0
lim
h
1
[
cos(x+h)
−2sin(
2
2x+h
)sin(−
2
h
)
]+4[−cosx
h→0
lim
h
(1−cosh)
−sinx
h→0
lim
h
sinh
]
=
cosx
5
⋅
h→0
lim
⎣
⎢
⎢
⎡
cos(x+h)
sin(
2
2x+h
)⋅
2
h
sin(
2
h
)
⎦
⎥
⎥
⎤
+4[(−cosx)⋅(0)−(sinx)⋅1]
=
cosx
5
⋅[
h→0
lim
cos(x+h)
sin(
2
2x+h
)
⋅
h→0
lim
2
h
sin(
2
h
)
]−4sinx
=
cosx
5
⋅
cosx
sinx
⋅1−4sinx
= 5secxtanx⋅−4sinx
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