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Differentiate Sin3xcos5x. .....w.r.t.x???
Answers
Sin 3x × Cos 5x =
1/2 { Sin ( 3x + 5x ) + Sin ( 3x - 5x ) }
= 1/2 { Sin 8x - Sin 2x }
Now, Differentiate W.r.t x we have
= 1/2 { 8 × Cos 8x - 2 × Cos 2x }
= { 4 Cos 8x - Cos 2x }
Answer:
y
′
=
3
cos
(
3
x
)
⋅
cos
(
5
x
)
−
5
sin
(
3
x
)
⋅
sin
(
5
x
)
Explanation:
First, notice that you can write your function as a product of two functions,
f
(
x
)
=
sin
(
3
x
)
and
g
(
x
)
=
cos
(
5
x
)
y
=
f
(
x
)
⋅
g
(
x
)
The product rule allows you to differentiate such functions by using the formula
d
d
x
(
y
)
=
[
d
d
x
(
f
(
x
)
)
]
⋅
g
(
x
)
+
f
(
x
)
⋅
d
d
x
(
g
(
x
)
)
In your case, you have
d
d
x
(
y
)
=
[
d
d
x
(
sin
(
3
x
)
)
]
⋅
cos
(
5
x
)
+
sin
(
3
x
)
⋅
d
d
x
(
cos
(
5
x
)
)
Now, in order to differentiate these two functions, you need to use the chain rule in the form
sin
u
1
and
cos
u
2
, with
u
1
=
3
x
and
u
2
=
5
x
.
This will get you
d
d
x
(
sin
u
1
)
−
[
d
d
u
1
sin
u
1
]
⋅
d
d
x
(
u
1
)
d
d
x
(
sin
u
1
)
=
cos
u
1
⋅
d
d
x
(
3
x
)
d
d
x
(
sin
(
3
x
)
)
=
cos
(
3
x
)
⋅
3
and
d
d
x
(
cos
u
2
)
=
[
d
d
u
2
cos
u
2
]
⋅
d
d
x
(
u
2
)
d
d
x
(
cos
(
5
x
)
)
=
−
sin
(
5
x
)
⋅
5
Your target derivative will thus be
y
′
=
[
3
⋅
cos
(
3
x
)
]
⋅
cos
(
5
x
)
+
sin
(
3
x
)
⋅
[
−
5
sin
(
5
x
)
]
y
′
=
3
cos
(
3
x
)
⋅
cos
(
5
x
)
−
5
sin
(
3
x
)
⋅
sin
(
5
x
)