]>
Theorem: For any positive integer
n
we have
ⅆ
ⅆ
x
(
x
n
)
=
n
⁢
x
n
−
1
.
Proof: Let
n
be a positive integer.
By defnition
ⅆ
ⅆ
x
(
x
n
)
=
lim
h
→
0
(
x
+
h
)
n
−
x
n
h
=
lim
h
→
0
(
x
n
+
n
⁢
x
n
−
1
⁢
h
+
terms with factor
⁢
h
2
)
−
x
n
h
=
lim
h
→
0
n
⁢
x
n
−
1
⁢
h
+
terms with factor
⁢
h
2
h
=
lim
h
→
0
h
⁡
(
n
⁢
x
n
−
1
+
terms with factor
⁢
h
)
h
=
lim
h
→
0
(
n
⁢
x
n
−
1
+
terms with factor
h
)
=
n
⁢
x
n
−
1
+
terms multiplied by
0
=
n
⁢
x
n
−
1