7. Differentiation Rules
P. K. Lamm
09/27/11 (17:33)
p. 1 / 6
Lecture Notes:
7. Differentiation Rules
These classnotes are intended to be supplementary to the textbook and are necessarily limited
by the time allotted for classes. For full and precise statements of definitions and theorems,
as well as material covering other topics and examples, please consult the textbook.
1. Rules of Differentiation
After some experience with computing derivatives using the definition of the derivative, the next
theorems may come as some relief.
Theorem:
If
f
(
x
) =
k
, for any constant
k
and all
x
, then
f
0
(
x
) = 0.
Proof:
Using the definition of the derivative,
f
0
(
x
)
=
lim
h
→
0
f
(
x
+
h
)

f
(
x
)
h
=
lim
h
→
0
k

k
h
=
0
.
Theorem:
If
n >
0 is an integer and
f
(
x
) =
x
n
, then
f
0
(
x
) =
nx
n

1
.
Proof:
The proof is not hard for cases
n
= 1 and
n
= 2, which we show first below:
For
n
= 1,
f
(
x
) =
x
, and
f
0
(
x
) = lim
h
→
0
f
(
x
+
h
)

f
(
x
)
h
= lim
h
→
0
(
x
+
h
)

x
h
= lim
h
→
0
h
h
= 1
= 1
·
x
0
.
For
n
= 2,
f
(
x
) =
x
2
, and
f
0
(
x
)
=
lim
h
→
0
(
x
+
h
)
2

x
2
h
=
lim
h
→
0
x
2
+ 2
xh
+
h
2

x
2
h
=
lim
h
→
0
h
(2
x
+
h
)
h
=
lim
h
→
0
2
x
+
h
=
2
x
= 2
x
1
.
1
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7. Differentiation Rules
P. K. Lamm
09/27/11 (17:33)
p. 2 / 6
For general
n
and
f
(
x
) =
x
n
, we use the Binomial Formula to write
(
x
+
h
)
n
=
x
n
+
nx
n

1
h
+
n
(
n

1)
2
x
n

2
h
2
+
· · ·
+
nxh
n

1
+
h
n
.
Then
f
0
(
x
)
=
lim
h
→
0
(
x
+
h
)
n

x
n
h
=
lim
h
→
0
x
n
+
nx
n

1
h
+
n
(
n

1)
2
x
n

2
h
2
+
· · ·
+
nxh
n

1
+
h
n

x
n
h
=
lim
h
→
0
h
nx
n

1
+
n
(
n

1)
2
x
n

2
h
+
· · ·
+
nxh
n

2
+
h
n

1
h
=
lim
h
→
0
nx
n

1
+
n
(
n

1)
2
x
n

2
h
+
· · ·
+
nxh
n

2
+
h
n

1
!
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 Fall '10
 KIHYUNHYUN
 Calculus, Derivative, lim, P. K. Lamm

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