Differentiation of sin2xcos3x
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Answered by
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d
d
x
[
f
(
x
)
g
(
x
)
]
is
f
(
x
)
d
d
x
[
g
(
x
)
]
+
g
(
x
)
d
d
x
[
f
(
x
)
]
where
f
(
x
)
=
sin
(
2
x
)
and
g
(
x
)
=
cos
(
3
x
)
.
sin
(
2
x
)
d
d
x
[
cos
(
3
x
)
]
+
cos
(
3
x
)
d
d
x
[
sin
(
2
x
)
]
Differentiate using the chain rule, which states that
d
d
x
[
f
(
g
(
x
)
)
]
is
f
'
(
g
(
x
)
)
g
'
(
x
)
where
f
(
x
)
=
cos
(
x
)
and
g
(
x
)
=
3
x
.
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sin
(
2
x
)
(
−
sin
(
3
x
)
d
d
x
[
3
x
]
)
+
cos
(
3
x
)
d
d
x
[
sin
(
2
x
)
]
Differentiate.
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−
3
sin
(
2
x
)
sin
(
3
x
)
+
cos
(
3
x
)
d
d
x
[
sin
(
2
x
)
]
Differentiate using the chain rule, which states that
d
d
x
[
f
(
g
(
x
)
)
]
is
f
'
(
g
(
x
)
)
g
'
(
x
)
where
f
(
x
)
=
sin
(
x
)
and
g
(
x
)
=
2
x
.
Tap for more steps...
−
3
sin
(
2
x
)
sin
(
3
x
)
+
cos
(
3
x
)
(
cos
(
2
x
)
d
d
x
[
2
x
]
)
Differentiate.
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−
3
sin
(
2
x
)
sin
(
3
x
)
+
2
cos
(
2
x
)
cos
(
3
x
)
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