differentiate y=sin^(3)x+cos^(3)x
Answers
Answered by
1
Step-by-step explanation:
y
=
sin
3
x
+
cos
3
x
d
y
d
x
=
d
d
x
(
sin
3
x
)
+
d
d
x
(
cos
3
x
)
Now we'll be using the chain rule to get
d
y
d
x
...
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
So, here it goes...
d
d
u
(
u
3
)
=
3
u
2
=
3
sin
2
x
d
d
x
(
sin
x
)
=
cos
x
d
d
u
(
u
3
)
⋅
d
d
x
(
sin
x
)
=
3
sin
2
x
cos
x
Which means that:
d
d
x
(
sin
3
x
)
=
3
sin
2
x
cos
x
Now...
d
d
p
(
p
3
)
=
3
p
2
=
3
cos
2
x
d
d
x
(
cos
x
)
=
−
sin
x
d
d
p
(
p
3
)
⋅
d
d
x
(
cos
x
)
=
−
3
cos
2
x
sin
x
Which means that...
d
d
x
(
cos
3
x
)
=
−
3
cos
2
x
sin
x
Since...
d
y
d
x
=
d
d
x
(
sin
3
x
)
+
d
d
x
(
cos
3
x
)
d
y
d
x
=
3
sin
2
x
cos
x
−
3
cos
2
x
sin
x
=
3
sin
x
cos
x
(
sin
x
−
cos
x
)
You can then again transform this result if you wish to.
Answered by
0
Step-by-step explanation:
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