Graphics Reference
In-Depth Information
)
¢
()
=
(
(
)( )( )
f
o
g
0
D f
o
g
0 1 ,
and use the fact that D(f g)(0) is a linear map. The details are left as Exercise
8.4.4.
Part (2) says that the map Df
p
generalizes the ordinary definition of the derivative
as defined in Section 4.3. Given a map f :
R
n
Æ
R
k
, we defined a derivative Df(
p
). If
we think of
R
n
and
R
k
as differentiable manifolds, then we just defined a new deriv-
ative Df
p
. The two derivatives are the same linear maps because of the definitions
involved and the fact that the tangent spaces at points of
R
n
o
and
R
k
are just
R
n
and
R
k
, respectively (Theorem 8.4.3(2)).
8.4.10. Example.
To compute Df
x
at a point x Œ
R
for the map
f:
RR
Æ
defined by
()
=
2
.
fx
x
Solution.
Let
v
ΠT
x
(
R
) =
R
. The curve
(
)
Æ
()
=+
g
:
-
11
,
R
,
g
txt
v
,
satisfies g(0) = x and g¢(0) =
v
. But
2
,
(
)( )
=+
(
)
f
o
g
t
xt
v
so that
)
¢
()
=+
)
¢
()
=
(
(
)
(
f
o
g
t
2
x
t
vv
and
f
o
g
0
2
x
v
.
We leave it to the reader to check that this agrees with Df(x)(
v
). What would have hap-
pened if we had chosen the curve
(
)
()
=+ +
3
h t
xt
t
v
,
which also satisfies h(0) = x and h¢(0) =
v
? Well,
2
[
(
)
]
3
(
)( )
=+ +
f
o
h
t
xt
t
v
and
)
¢
()
=++
[
(
)
]
(
)
(
3
2
f
o
h
t
2
x
t
t
v
3
t
+
1
v
,
)
¢
()
02
v
,
(
so that
f
o
h
x
which agrees with the answer we got using the curve g(t).
