I-Calculus
6 - Power Rule for Differentiation (5)
Let
y
=
f
(
x
)
=
x
m
n
,where
m
is a positive integer and
n
is any integer. Then
d
y
d
x
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
(
x
+
h
)
m
n
−
x
m
n
h
=
lim
h
→
0
(
(
x
+
h
)
1
n
)
m
−
(
x
1
n
)
m
h
=
lim
h
→
0
[
(
x
+
h
)
1
n
−
x
1
n
]
[
(
(
x
+
h
)
1
n
)
m
−
1
+
(
(
x
+
h
)
1
n
)
m
−
2
x
1
n
+
⋅
⋅
⋅
+
(
x
+
h
)
1
n
(
x
1
n
)
m
−
2
+
(
x
1
n
)
m
−
1
]
h
=
lim
h
→
0
(
x
+
h
)
1
n
−
x
1
n
h
[
(
(
x
+
h
)
1
n
)
m
−
1
+
(
(
x
+
h
)
1
n
)
m
−
2
x
1
n
+
⋅
⋅
⋅
+
(
x
+
h
)
1
n
(
x
1
n
)
m
−
2
+
(
x
1
n
)
m
−
1
]
=
1
n
x
1
n
−
1
[
m
x
m
−
1
n
]
=
m
n
x
m
n
−
1
Newer Post
Older Post
Home