Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

204 D CHAPTER 14 PARTIAL DERIVATIVES
tangent plane become almost indistinguishable. (Here, the tangent plane is below the surface.) If we zoom in farther, the
surface and the tangent plane will appear to coincide.
_ xy sin {X - y)
9 · !( x,y ) - 2 2
1+x +y
ACAS . f ( ' ) ysin(x - y) + xycos(x-y) 2x2 ysin{x- y) d
g1ves "' x,y = - an
1 + x2 + y2 (1 + x2 + y2)2
fv(x, y) =
x sin ( x - y) - xy cos ( x - y) 2xy 2 sin ( x - y) ·
- ( 2 . We use the CAS to evaluate these at (1 , 1), and then
+ x + y 1 + x2 + y2)
1
2 2
substitute the results into Equation 2 to compute an equation oqhe tangent plane: z = tx - tY· The surface and tangent
plane are shown in the first graph below. After zooming in, the surface and the tangent plane become almost indistinguishable,
as shown in the second graph. (Here, the tangent plane is shown with fewer traces than the surface.) If we zoom in farther, the
surface and the tangent plane will appear to coincide.
z 0
-1
11. f(x, y) = 1 + x Jn(xy - 5): The partlal derivatives are f, (x, y) = x · -
and fv(x, y) = x · -
xy-5 xy- ,
1 - (y) + Jn (xy - 5) · 1 = xy 5
+ln(xy- 5)
. xy - 5 xy -
1 - (x) = __£_ 5
, so f x(2 , 3) = 6 and / 11 (2, 3) = 4. Both f , and / 11 are continuous functions for
xy > 5, so by Theorem 8, f is differentiable at (2, 3). By Equation 3, the linearization off at (2, 3) is given· by
L(x, y) = /(2, 3) + f x (2, 3)(x- 2) + f v (2, 3)(y - 3) = 1 + 6(x - 2) + 4(y - 3) = 6x + 4y - 23.
13. f(x, y) ~ _ x _ . The partial !;lerivatives are f x(x, y) = l(x t y) ) ~'t( 1 ) = y/ (x + y) 2 and
x + y x +y
fv(x, y) = x ( -1){x + y)- 2 • 1 = -x/ (x + y) 2 , so f x(2, 1) = ~and / 11
(2; 1) = - ~.
Both f, and j 11 are continuous
functions for y =I - x, so f is differentiable at (2, 1) by Theorem 8. The linearization of f at (2, 1) is given by
L(x, y) = !(2, 1) + f x(2 , 1)(x - 2) + / 11 (2;1)(y - 1) = 1 + ~ (x - 2) - ~(y - 1) = ~ X - ~ y + ~·
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