James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

938 Chapter 14 Partial Derivatives
is differentiable (Definition 14.4.7). Recall that this is the case when f x and f y are continuous
(Theorem 14.4.8).
2 The Chain Rule (Case 1) Suppose that z − f sx, yd is a differentiable function
of x and y, where x − tstd and y − hstd are both differentiable functions of t.
Then z is a differentiable function of t and
dz
dt − −f
−x
dx
dt 1 −f
−y
dy
dt
Proof A change of Dt in t produces changes of Dx in x and Dy in y. These, in turn,
produce a change of Dz in z, and from Definition 14.4.7 we have
Dz − −f −f
Dx 1
−x −y Dy 1 « 1 Dx 1 « 2 Dy
where « 1 l 0 and « 2 l 0 as sDx, Dyd l s0, 0d. [If the functions « 1 and « 2 are not
defined at s0, 0d, we can define them to be 0 there.] Dividing both sides of this equation
by Dt, we have
Dz
Dt − −f
−x
Dx
Dt 1 −f
−y
Dy
Dt 1 « Dx
1
Dt 1 « Dy
2
Dt
If we now let Dt l 0, then Dx − tst 1 Dtd 2 tstd l 0 because t is differentiable and
therefore continuous. Similarly, Dy l 0. This, in turn, means that « 1 l 0 and « 2 l 0, so
dz
dt − lim Dz
Dt l 0 Dt
− −f
−x lim Dx
Dt l 0 Dt 1 −f
−y lim Dy
Dt l 0 Dt 1 a lim « Dx
1b lim
Dt l 0 Dt l 0 Dt 1 a lim « Dy
2b lim
Dt l 0 Dt l 0 Dt
− −f
−x
dx
dt 1 −f
−y
dy
dt 1 0 ? dx
dt 1 0 ? dy
dt
− −f
−x
dx
dt 1 −f
−y
dy
dt
■
Notice the similarity to the definition of
the differential:
dz − −z −z
dx 1
−x −y dy
Since we often write −zy−x in place of −fy−x, we can rewrite the Chain Rule in the form
dz
dt − −z
−x
dx
dt 1 −z
−y
dy
dt
EXAMPLE 1 If z − x 2 y 1 3xy 4 , where x − sin 2t and y − cos t, find dzydt when
t − 0.
SOLUTION The Chain Rule gives
dz
dt − −z
−x
dx
dt 1 −z
−y
dy
dt
− s2xy 1 3y 4 ds2 cos 2td 1 sx 2 1 12xy 3 ds2sin td
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