Math 23
Sections 110113
B. Dodson
Week 6 Homework:
14.3 partial derivatives, 2nd order deriv
14.4 tangent plane, diﬀerentials
14.5 chain rule
Problem 14.3.15:
Find the partial derivatives of
the function
z
=
f
(
x, y
) =
xe
3
y
.
Find
f
x
(2
,
1)
.
Solution:
For
f
x
we (temporarily) hold
y
constant,
so
e
3
y
constant, giving
f
as constant
·
x
.
We then take the derivative in
x
, with
(constant
·
x
)’ = constant, or
f
x
=
e
3
y
.
Likewise, with
x
constant,
f
y
=
x
∂
∂y
(
e
3
y
)
=
x
d
dy
(
e
3
y
) = 3
xe
3
y
.
Finally,
f
x
(2
,
1) =
e
3
gives the rate of change
of
f
at (2
,
1) with respect to
x
; which we
may constrast with
f
y
(2
,
1) = 6
e
3
,
the rate of
change of
f
at (2
,
1) with respect to
y
.
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.
For example,
f
is growing six times more rapidly
in
y
than in
x
. We also solved #49, 14.3, and
in particular veriﬁed that the second partials
z
xy
=
z
yx
.
Week 6 Homework:
14.4 tangent plane, diﬀerentials
14.5 chain rule
Problem 14.4.3:
Find the tangent plane to the surface
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 Spring '06
 YUKICH
 Chain Rule, Derivative

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