I-Calculus
2 - Power Rule for Differentiation (1)
Let
y
=
f
(
x
)
=
x
n
, where
n
is a positive integer. Then
d
y
d
x
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
(
x
+
h
)
n
−
x
n
h
=
lim
h
→
0
[
(
x
+
h
)
−
x
]
[
(
x
+
h
)
n
−
1
+
(
x
+
h
)
n
−
2
x
+
(
x
+
h
)
n
−
3
x
2
+
⋅
⋅
⋅
+
(
x
+
h
)
x
n
−
2
+
x
n
−
1
]
h
=
lim
h
→
0
h
[
(
x
+
h
)
n
−
1
+
(
x
+
h
)
n
−
2
x
+
(
x
+
h
)
n
−
3
x
2
+
⋅
⋅
⋅
+
(
x
+
h
)
x
n
−
2
+
x
n
−
1
]
h
=
lim
h
→
0
[
(
x
+
h
)
n
−
1
+
(
x
+
h
)
n
−
2
x
+
(
x
+
h
)
n
−
3
x
2
+
⋅
⋅
⋅
+
(
x
+
h
)
x
n
−
2
+
x
n
−
1
]
=
n
x
n
−
1
If
y
=
f
(
x
)
=
k
, where
k
is a constant, then
d
y
d
x
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
k
−
k
h
=
lim
h
→
0
0
h
=
lim
h
→
0
0
=
0
In particular, if
y
=
f
(
x
)
=
1
=
x
0
, then
d
y
d
x
=
0
=
0
x
0
−
1
Thus we see that if
y
=
f
(
x
)
=
x
n
, where
n
is a nonnegative integer, then
d
y
d
x
=
n
x
n
−
1
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