Solução_Calculo_Stewart_6e

F.
180 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
47. x − z =arctan(yz) ⇒ ∂
∂x
1=
∂
∂z
(x − z) = (arctan(yz)) ⇒ 1 −
∂x ∂x = 1
1+(yz) · y ∂z
2 ∂x
y ∂z
y +1+y 2
1+y 2 z +1 2 ∂x ⇔ 1= z 2 ∂z
1+y 2 z 2 ∂x ,so∂z ∂x = 1+y2 z 2
1+y + y 2 z . 2
∂
∂y (x − z) = ∂
∂z
(arctan(yz)) ⇒ 0 −
∂y ∂y = 1
1+(yz) · y ∂z
2 ∂y + z · 1
z
−
1+y 2 z = y ∂z
2 1+y 2 z +1 z y +1+y 2
⇔ −
2 ∂y 1+y 2 z = z 2 ∂z
2 1+y 2 z 2 ∂y
49. (a) z = f(x)+g(y) ⇒ ∂z
∂x = f 0 (x),
(b) z = f(x + y).
∂z
∂y = g0 (y)
Let u = x + y. Then ∂z
∂x = df ∂u
du ∂x = df
du (1) = f 0 (u) =f 0 (x + y),
∂z
∂y = df ∂u
du ∂y = df
du (1) = f 0 (u) =f 0 (x + y).
⇔
⇔
⇔
∂z
∂y = − z
1+y + y 2 z . 2
51. f(x, y) =x 3 y 5 +2x 4 y ⇒ f x(x, y) =3x 2 y 5 +8x 3 y, f y(x, y) =5x 3 y 4 +2x 4 .Thenf xx(x, y) =6xy 5 +24x 2 y,
f xy (x, y) =15x 2 y 4 +8x 3 , f yx (x, y) =15x 2 y 4 +8x 3 ,andf yy (x, y) =20x 3 y 3 .
53. w = √ u 2 + v 2 ⇒ w u = 1 2 (u2 + v 2 ) −1/2 · 2u =
u
√
u2 + v , w 2 v = 1 2 (u2 + v 2 ) −1/2 v
· 2v = √
u2 + v .Then 2
w uu = 1 · √u 2 + v 2 − u · 1
2 (u2 + v 2 ) −1/2 √
(2u) u2 + v
√
u2 + v 2 2
=
2 − u 2 / √ u 2 + v 2
= u2 + v 2 − u 2
u 2 + v 2 (u 2 + v 2 ) = v 2
3/2 (u 2 + v 2 ) , 3/2
w uv = u
− 1 2 u 2 + v 2−3/2 uv
(2v) =−
(u 2 + v 2 ) , w 3/2 vu = v
− 1 2 u 2 + v 2−3/2 uv
(2u) =−
(u 2 + v 2 ) , 3/2
w vv = 1 · √u 2 + v 2 − v · 1
2 (u2 + v 2 ) −1/2 √
(2v) u2 + v
√
u2 + v 2 2
=
2 − v 2 / √ u 2 + v 2
= u2 + v 2 − v 2
u 2 + v 2 (u 2 + v 2 ) = u 2
3/2 (u 2 + v 2 ) . 3/2
55. z =arctan x + y
1 − xy
⇒
z x =
1+
=
z y =
1+
1
x+y
1−xy
2 ·
1+y 2
(1 + x 2 )(1 + y 2 ) = 1
1+x 2 ,
1
x+y
1−xy
2 ·
(1)(1 − xy) − (x + y)(−y) 1+y 2
=
(1 − xy) 2 (1 − xy) 2 +(x + y) = 1+y 2
2 1+x 2 + y 2 + x 2 y 2
(1)(1 − xy) − (x + y)(−x) 1+x 2
=
(1 − xy) 2 (1 − xy) 2 +(x + y) = 1+x 2
2 (1 + x 2 )(1 + y 2 ) = 1
1+y . 2
Then z xx = −(1 + x 2 ) −2 2x
· 2x = −
(1 + x 2 ) , 2 zxy =0, zyx =0, zyy = −(1 + y2 ) −2 2y
· 2y = −
(1 + y 2 ) . 2
57. u = x sin(x +2y) ⇒ u x = x · cos(x +2y)(1) + sin(x +2y) · 1=x cos(x +2y)+sin(x +2y),
u xy = x(− sin(x +2y)(2)) + cos(x +2y)(2) = 2 cos(x +2y) − 2x sin(x +2y),
u y = x cos (x +2y)(2)=2x cos(x +2y),
u yx =2x · (− sin(x +2y)(1)) + cos (x +2y) · 2=2cos(x +2y) − 2x sin(x +2y). Thus u xy = u yx .