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

198 D CHAPTER 14 PARTIAL DERIVATIVES
25. g(u,v) = (u 2 v- v 3 ) 5 =? gu(u,v) = 5(u 2 v - v 3 ) 4 • 2uv = 10u·u(u 2 v - v 3 )\
gu(u, v) = 5(u 2 v- v 3 ) 4 (u 2 - 3v 2 ) = 5(u 2 - 3v 2 )(u 2 v- v 3 ) 4
. '
29. F(x,y) = l " cos(et)dt =? Fx(x,y)= :xl" cos (e 1 ) dt=cos(e"' ) by theFundamentaiTheoremofCalculus, Partl;
F 11 (x,y) = o·y
0 1"' . t
11
cos(e) dt = oy - I. cos(e) dt = - oy f .. cos(et) dt = - cos(e 11 ).
0 [ ( 11 . t ] 0 rv
33. w = ln(x + 2y + 3z)
ow 1 ow 2 ow 3
=> ox = x + 2y + 3z' oy = x + 2y + 3z' oz = x + 2y + 3z
35. u = xysin- ~(yz) =>
ou 1 xy 2
- = xy . (y) = --;:::.:=:::::::::==:::
az J 1- (yz)2 J 1- y2z2
37. h(x, y, z, t) = x 2 y cos(z/t) => h.,(x; y, z, t) = 2xy cos(z/t), h 11 (x, y, z, t) = x 2 cos(z/t),
hz(x, y, z, t) = -x 2 y sin(z/t)(1/ t) = ( - x 2 y/t) sin(z/t), ht(x, y, z, t) = -x 2 y sin(z/t)( -zC 2 ) = (x 2 yz/t 2 ) sin.(z/t)
39 V 2 2 2 F h · 1
1 ( 2 2 2) -1/2 ( )
. u = x
Xi
1 + x 2 + ··· + Xn· or eac t = , ..., n, u,, = 2 x 1 + X2 + ·· · + Xn 2x; = .
y X~ + X~ + ··· + x?,
41. f(x, y) = 1n( x + Jx2 + y2) '*
( ) _ 1 [ 1 ( 2 2)- 1/ 2( )] _ 1 ( X )
f., x,y - ~ 1+2 x +y 2x - ~ 1+ ~ ,
x + Y x2 + y2 x + Y x2 + y2 Y x2 + y2
. 1 ( 3 ) .
so f.,(, 34= ) 1 =11 !!=l
3
+~ + ...f32+42 s ( + s) s· .
y l(x + y + z)- y(l ) x + z
43. f(x,y, z) = x+y + z =? fv(x,y,z) = (x + y + z )2 = (x+y+ z )2'
2+(-1) 1
soj11(2,1, - 1) = ( 2 + 1 +(- 1 )) 2 4·
45. f(x, y) = x1l - x 3 y =>
f ( ) _ 1 . f(x + h,y) - f(x,y) _ 1 . (x +h)y2 - (x + h) 3 y- (xy 2 - x 3 y)
"' x, y - liD - liD
,...... o h h-0 h .
h(y 2 - 3x 2 y - 3xyh- yh 2 )
= lim = lim (y 2 - 3x 2 y- 3xyh- yh 2 ) = y 2 - 3x 2 y
,. ..... o h h ..... o
f ( ) _ li f(x, y + h) - f(x;y) _ U:U x(y + h) 2 - x 3 (y + h)- (xy 2 - x 3 y) _ 1' h(2xy + xh- x 3 )
11 x , y - h5 h - h ->0 h - "~ h
· = lim {2xy + xh- x 3 ) = 2xy - x 3
/a.-+0 .
© 2012 Ccngnge Learning. All Rights Reserved. Mny not be scanned, copied, orduplicoted, or posted to • publicly ncc=iblc website, in whole or in P."t.