For next Wednesday, read section 2.3.
Last time we looked at the function
f
(
x
) =
x

x
:

2
 1
1
2
 4
 2
2
4
which can also be described by the twopart formula
{–
x
2
if
x
< 0,
f
(
x
)
=
{
{
x
2
if
x
≥
0.
We showed last time that
f
(
x
) is differentiable at
x
=0:
lim
h
→
0+
(
h

h
 – 0)/
h
= lim
h
→
0+
h
2
/
h
= lim
h
→
0+
h
= 0 and
lim
h
→
0–
(
h

h
 – 0)/
h
= lim
h
→
0–
–
h
2
/
h
= lim
h
→
0–
–
h
= 0.
Since the lefthand and righthand limits agree, the two
sided limit exists.
So the derivative of
f
(
x
) at
x
=0 is 0.
What is the derivative of
f
(
x
) =
x

x
 at
x
=1? ...
..?..
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Note that for all
h
sufficiently close to 0 (specifically, for
all
h
with 
h
 < 1),
f
(1+
h
) = (1+
h
)
2
.
Hence the twosided
limit lim
h
→
0
(
f
(1+
h
) –
f
(1))/
h
equals the twosided limit
lim
h
→
0
((1+
h
)
2
– (1)
2
)/
h
= lim
h
→
0
(2 +
h
) = 2.
[Explain this with a picture of
y
=
x
2
superimposed on a
picture of
y
=
x

x
, and various secant lines.]
The general principle at work here is:
Locality principle for derivatives:
If there exists some
r
> 0 such that
f
(
x
) =
g
(
x
) for all
x
in
(
a
–
r
,
a
+
r
), then
f
′
(
a
) =
g
′
(
a
).
[Draw a picture.]
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 Fall '11
 Staff
 Calculus, Derivative, locality principle

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