Math 136, Answers to Review Problems for Test 1
1. We calculate
D
f
using the chain rule and the fact that
F
(
x, y, f
(
x, y
)) = 0. We define
G
(
x, y
) =
x
y
f
(
x, y
)
and note that
D
g
=
1
0
0
1
∂f
∂x
∂f
∂y
.
We know
G
is differentiable
since its coordinate functions are
C
1
(the first two coordinate functions are
C
∞
, and
the third,
f
, is
C
1
since
F
and (
x
0
, y
0
, z
0
) satisfy the Implicit Function Theorem).
Using the chain rule on the equation
F
(
x, y, f
(
x, y
)) = 0, we get
0
=
D
0 =
D
(
F
(
x, y, f
(
x, y
)
)
=
D
(
F
◦
G
) =
D
F
·
D
G
where
·
is matrix multiplication. Now, using the fact that
D
F
(
x, y, z
) =
(
∂F
∂x
,
∂F
∂y
,
∂F
∂z
)
we see
(1)
0 =
∂F
∂x
+
∂F
∂z
∂f
∂x
,
∂F
∂y
+
∂f
∂y
∂F
∂z
.
We can solve (1) for
∂f
∂x
and
∂f
∂y
and this give
∂f
∂x
=

∂F
∂x
∂F
∂z
and
∂f
∂y
=

∂F
∂y
∂F
∂z
evaluated
at (
x
0
, y
0
, f
(
x
0
, y
0
)). Here we used that
∂F
∂z
(
x
0
, y
0
, z
0
) = 0 (and
z
0
=
f
(
x
0
, y
0
)).
2. The function
F
in this problem satisfies the hypotheses of the Implicit Function Theo
rem, so we can assert there is an open neighborhood
U
of
x
0
and an open neighborhood
V
of
y
0
and a
C
1
function
f
:
U
→
V
such that
F
(
x, f
(
x
)) =
F
(
x
0
, y
0
) for all
x
∈
U
.
(a) So, as there are an infinite number of points in
U
, and for each
x
∈
U
,
F
(
x, f
(
x
)) =
F
(
x
0
, y
0
), there are an infinite number of points that satisfy
F
(
x, y
) =
F
(
x
0
, y
0
).
(b) No by (a).
(c) This might be proven in class, but if not, here is the proof. Let
G
be the auxiliary
function in the Implicit Function Theorem,
G
(
x, y
) = (
x, F
(
x, y
)). In the proof