Graphics Reference
In-Depth Information
4.3.7. Corollary.
Let
U
be an open set in
R
n
and let f, g :
U
Æ
R
be functions that
are differentiable at
p
Œ
U
.
(1) D(f + g)(
p
) = Df(
p
) + Dg(
p
)
(2) D(fg)(
p
) = f(
p
)Dg(
p
) + g(
p
)Df(
p
)
(3) If g(
p
) π 0, then
() ()
-
() ()
()
gDf
pp p p
p
2
fDg
(
)( )
=
Df g
p
.
g
Proof.
We prove (1) to show how the chain rule gets used. The rest are left as an
exercise. Define functions m(
p
) = (f(
p
),g(
p
)) and s(
p
,
q
) =
p
+
q
. Clearly, (f + g)(
p
) =
s(m(
p
)). Therefore, the chain rule, Proposition 4.3.4(2), and Proposition 4.3.5 and
imply that
(
)( )
=
(
)( )
=
(
()
)
( )
=
(
()
()
)
=
()
+
()
D f
+
g
p
D
sm
o
p
D
sm
p
o
D
m
p
s
o
Df
p
,
Dg
p
Df
p
Dg
p
.
Let f(u) = (sin u,u
2
), g(x,y) = x
2
4.3.8. Example.
+ 3xy, and G(u) = g(f(u)). Find the
derivative DG.
Solution.
In this problem, it is of course easy to compose the functions f and g to
get
()
=
2
2
Gu
sin
u
+
3
u
sin
u
and
¢
()
=
2
Gu
2
sin
u
cos
u
+
3
u
cos
u
+
6
u
sin
u
.
On the other hand, using the chain rule we do not need to compute G(u) directly. Note
that
()()
=
(
(
)
(
)
)
Dfa h
cos
ah
, 2
ah
and
(
)(
)
=+ +
(
)
Dg c d h k
,
,
2
ch
3
dh
3
ck
.
It follows that
(
)
(
()()
=
2
(
)
(
)
)
DG a h
Dg
sin ,
a a a h a h
aaha ahaah
cos
,
2
+
(
)
+
(
=
(
)
2
)
2
sin
cos
3
cos
6
sin
,
which agrees with the first answer.
Although the chain rule lets us compute the derivatives of many functions, its
direct use is rather messy. To make computations still easier, we use the chain rule to
determine the matrix for the linear transformation Df(
p
).
