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

SectioN 14.5 The Chain Rule 943
Now we suppose that z is given implicitly as a function z − f sx, yd by an equation of
the form Fsx, y, zd − 0. This means that Fsx, y, f sx, ydd − 0 for all sx, yd in the domain
of f . If F and f are differentiable, then we can use the Chain Rule to differentiate the
equation Fsx, y, zd − 0 as follows:
But
−F
−x
−x
−x 1 −F
−y
−y
−x 1 −F
−z
−z
−x − 0
−
−x sxd − 1 and −
−x syd − 0
so this equation becomes
−F
−x 1 −F
−z
−z
−x − 0
If −Fy−z ± 0, we solve for −zy−x and obtain the first formula in Equations 7. The formula
for −zy−y is obtained in a similar manner.
7
−z
−x − 2
−F
−x
−F
−z
−z
−y − 2
−F
−y
−F
−z
The solution to Example 9 should be
compared to the one in Example 14.3.5.
Again, a version of the Implicit Function Theorem stipulates conditions under which
our assumption is valid: if F is defined within a sphere containing sa, b, cd, where
Fsa, b, cd − 0, F z sa, b, cd ± 0, and F x , F y , and F z are continuous inside the sphere, then
the equation Fsx, y, zd − 0 defines z as a function of x and y near the point sa, b, cd and
this function is differentiable, with partial derivatives given by (7).
EXAMPLE 9 Find −z −z
and
−x −y if x 3 1 y 3 1 z 3 1 6xyz − 1.
SOLUTION Let Fsx, y, zd − x 3 1 y 3 1 z 3 1 6xyz 2 1. Then, from Equations 7, we
have
−z
−x − 2 F x
F z
−z
−y − 2 F y
F z
− 2 3x 2 1 6yz
3z 2 1 6xy − 2 x 2 1 2yz
z 2 1 2xy
− 2 3y 2 1 6xz
3z 2 1 6xy − 2 y 2 1 2xz
z 2 1 2xy
■
1–6 Use the Chain Rule to find dzydt or dwydt.
1. z − xy 3 2 x 2 y, x − t 2 1 1, y − t 2 2 1
2. z − x 2 y
x 1 2y , x − et , y − e 2t
3. z − sin x cos y, x − st , y − 1yt
4. z − s1 1 xy , x − tan t, y − arctan t
5. w − xe yyz , x − t 2 , y − 1 2 t, z − 1 1 2t
6. w − lnsx 2 1 y 2 1 z 2 , x − sin t, y − cos t, z − tan t
7–12 Use the Chain Rule to find −zy−s and −zy−t.
7. z − sx 2 yd 5 , x − s 2 t, y − st 2
8. z − tan 21 sx 2 1 y 2 d, x − s ln t, y − te s
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