To Find:
Value of,
Solution:
W.k.t
By using the identities
#
Answer:
Let f(x)=cos(x−
8
π
)
Thus using first principle
f
′
(x)=
x→0
lim
h
f(x+h)−f(x)
=
1
[cos(x+h−
)−cos(x−
)]
[−2sin
2
(x+h−
+x−
sin(
x+h−
−x+
[−2sin(
2x+h−
4
)sin
]
[−sin(
(
=−sin(
2x+0−
)⋅1=−sin(x−
Step-by-step explanation:
PN MAJHI JAGA DUSRYA KUNALA DILAS TR BG TU
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Answers & Comments
To Find:
Value of,
Solution:
W.k.t
By using the identities
#![\texttt{\red{ Marzi} \: \blue{Here}...!} \texttt{\red{ Marzi} \: \blue{Here}...!}](https://tex.z-dn.net/?f=%5Ctexttt%7B%5Cred%7B%20Marzi%7D%20%5C%3A%20%5Cblue%7BHere%7D...%21%7D)
Answer:
Let f(x)=cos(x−
8
π
)
Thus using first principle
f
′
(x)=
x→0
lim
h
f(x+h)−f(x)
=
x→0
lim
h
1
[cos(x+h−
8
π
)−cos(x−
8
π
)]
=
x→0
lim
h
1
[−2sin
2
(x+h−
8
π
+x−
8
π
)
sin(
2
x+h−
8
π
−x+
8
π
)]
=
x→0
lim
h
1
[−2sin(
2
2x+h−
4
π
)sin
2
h
]
=
x→0
lim
[−sin(
2
2x+h−
4
π
)
(
2
h
)
sin(
2
h
)
]
=−sin(
2
2x+0−
4
π
)⋅1=−sin(x−
8
π
)
Step-by-step explanation:
PN MAJHI JAGA DUSRYA KUNALA DILAS TR BG TU