Section 11.5: The chain rule (concluded)
Implicit differentiation revisited: Suppose (
a
,
b
) is a point
on the curve
C
= {(
x
,
y
):
F
(
x
,
y
) = 0} (e.g., when the function
F
(
x
,
y
) is
x
2
+
y
2
– 1, the curve
C
is the unit circle), so that
F
(
a
,
b
) = 0.
We’d like to know that there exists a piece of the curve that
is the graph of a function and contains the point (
a
,
b
).
That is, we’d like to know that there exists a function
f
defined on some interval (
a
–
r
,
a
+
r
) (with
r
> 0) such that
f
(
a
) =
b
and
F
(
x
,
f
(
x
)) = 0 for all
x
in (
a
–
r
,
a
+
r
).
But how can we be sure that the equation
F
(
x
,
y
) = 0
determines
y
as a function of
x
in the vicinity of a particular
point (
a
,
b
) satisfying
F
(
a
,
b
) = 0?
In fact, we’ve seen that this doesn’t work for the points
(
±
1,0) on the circle
x
2
+
y
2
=1!
The Implicit Function Theorem tells us that the equation
F
(
x
,
y
) = 0 determines
y
as a function of
x
in the vicinity of
a particular point (
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 Fall '11
 Staff
 Chain Rule, Derivative, Implicit Differentiation, The Chain Rule, Unit Circle, implicit function theorem, gradient vector, directional derivatives, Du f

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