Differentiate ( using an embedded chain rule) : x ^2 (x+1) 3
Answers
Answer:
So because you have a fraction you need to use the quotient rule and the chain rule together.
The quotient rule is: (
(
f
g
)
'
=
f
'
g
−
g
'
f
g
2
.
The chain rule is: F'=
f
'
(
g
(
x
)
)
⋅
g
'
(
x
)
Therefore:
1
'
(
x
2
−
2
x
−
5
)
4
−
1
(
x
2
−
2
x
−
5
)
4
(
(
x
2
−
2
x
−
5
)
4
)
2
That gives you
0
−
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
x
2
−
2
x
−
5
)
8
0
−
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
(
x
2
−
2
x
−So because you have a fraction you need to use the quotient rule and the chain rule together.
The quotient rule is: (
(
f
g
)
'
=
f
'
g
−
g
'
f
g
2
.
The chain rule is: F'=
f
'
(
g
(
x
)
)
⋅
g
'
(
x
)
Therefore:
1
'
(
x
2
−
2
x
−
5
)
4
−
1
(
x
2
−
2
x
−
5
)
4
(
(
x
2
−
2
x
−
5
)
4
)
2
That gives you
0
−
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
x
2
−
2
x
−
5
)
8
0
−
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
(
x
2
−
2
x
−
5
)
4
)
2
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
x
2
−
2
x
−
5
)
8
Then take the expression in the parenthesis to the power of 3 and then distribute 4 and multiply by
2
x
−
2
Take the denominator to the power 8 and then cancel out top and bottom.
It could get very long and frustrating but make sure to avoid any small mistakes along the way.
Good luck
5
)
4
)
2
4
(
x
2
−
2
x
−
5
)
3
⋅
2
x
−
2
(
x
2
−
2
x
−
5
)
8
Then take the expression in the parenthesis to the power of 3 and then distribute 4 and multiply by
2
x
−
2
Take the denominator to the power 8 and then cancel out top and bottom.
It could get very long and frustrating but make sure to avoid any small mistakes along the way.
Good luck