The nth derivative of (x+1)^-2
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Answer:
Explanation:
Let us look at the first few derivatives to see what's happening:
f
(
0
)
(
x
)
=
x
1
2
f
(
1
)
(
x
)
=
1
2
x
−
1
2
f
(
2
)
(
x
)
=
−
1
4
x
−
3
2
f
(
3
)
(
x
)
=
3
8
x
−
5
2
f
(
4
)
(
x
)
=
−
15
16
x
−
7
2
f
(
5
)
(
x
)
=
105
32
x
−
9
2
f
(
6
)
(
x
)
=
−
945
64
x
−
11
2
The coefficient is a product of odd numbers divided by a power of
2
.
Note that:
1
=
1
⋅
2
2
=
2
!
2
1
⋅
1
!
1
⋅
3
=
1
⋅
2
⋅
3
⋅
4
2
⋅
4
=
4
!
2
2
⋅
2
!
1
⋅
3
⋅
5
=
1
⋅
2
⋅
3
⋅
4
⋅
5
⋅
6
2
⋅
4
⋅
6
=
6
!
2
3
⋅
3
!
etc.
So we can write:
f
(
2
)
(
x
)
=
(
−
1
)
2
−
1
2
!
2
3
⋅
1
!
x
−
3
2
f
(
3
)
(
x
)
=
(
−
1
)
3
−
1
4
!
2
5
⋅
2
!
x
−
5
2
f
(
4
)
(
x
)
=
(
−
1
)
4
−
1
6
!
2
7
⋅
3
!
x
−
7
2
f
(
5
)
(
x
)
=
(
−
1
)
5
−
1
8
!
2
9
⋅
4
!
x
−
9
2
So it looks like a valid formula for
n
>
1
would be:
f
(
n
)
(
x
)
=
(
−
1
)
n
−
1
(
2
n
−
2
)
!
2
2
n
−
1
(
n
−
1
)
!
x
1
−
2
n
2
If
n
=
1
then it gives us:
f
(
1
)
(
x
)
=
(
−
1
)
1
−
1
(
2
−
2
)
!
2
2
−
1
(
1
−
1
)
!
x
1
−
2
2
f
(
1
)
(
x
)
=
1
2
x
−
1
2
which is correct too.
So this formula seems to work for all
n
≥
1
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