problem set 2 answers (New !! on November 13)

Calculus 2 Problem Set 2 Answers October 2014
[1] (1)
定義域 {(x, y)|x2 + y 2 < 1}, 値域 {z|z ≥ 1}
(2) 定義域 {(x, y)|xy > 0}, 値域 {z| − ∞ < z < ∞}
[2]
√
1/2, (3) 2 6, (4) 1, (5)
√
√
√
√
x−y+2 x−2 y
√
√
= x + y + 2 を用いる。2,
x− y
(1) 5/2,
(7)
(2)
x2 −2xy+y 2
x−y
1, (6)
(8)
= x − y を用いる。0,
0
[3]
(1) y = mx に沿って近づけて極限を調べる, (2)
y = mx2 に沿って近づける。
[4]
(2) fx = 4xey , fy = 2x2 ey , (3)
(1) fx = 6x, fy = 2,
(4) fx = −y 2 /x2 , fy = 2y/x, (5)
fx = 2x + yexy , fy = xexy ,
fx = 1/(1 + ey ), fy = −xey /(1 + ey )2 ,
(6) fx = 12(9x2 y + 3x)11 (18xy + 3), fy = 12(9x2 y + 3x)11 (9x2 ),
(7) fx = e3x (2x log y + 3x2 log y), fy = x2 e3x /y,
(8) fx = (1 + xy − y log y)exy , fy = (x2 − x log y − 1/y)exy
(9) fx = 2y/(x + y)2 , fy = −2x/(x + y)2 , (10)
(11) fx = 4x + 3y, fy = 3x − 8y,
(12)
(14) zx = sin y, zy = x cos y, (15)
fx = 2y(1 − x)/ex , fy = 2x/ex
zx = y1 , zy = − yx2 ,
(18) fx =
y
−x2 +y 2
(x2 +y 2 )2 , fy
=
−2xy
(x2 +y 2 )2 ,
[5]
fx (2, −3) = 1, fy (2, −3) = 3
[6]
fx (1, 2) = 6, fy (1, 2) = −1
fx = x1 , fy = − y1
fx = cos x − sin(x + y), fy = cos y − sin(x + y),
(16) zx = e cos y + e cos x, zy = −ex sin y + ey sin x,
x
(13)
(17) zx = (x2 + 2x + 2y)ex , zy = 2ex ,
(19) zx = cos 2x, zy = cos 2y.
[7]
(1) fx = 3x2 y + 2y 2 , fxx = 6xy, fy = x3 + 4xy, fyy = 4x, fxy = fyx = 3x2 + 4y,
(2) fx = ey + 4x3 y, fxx = 12x2 y, fy = xey + x4 + 3y 2 , fyy = xey + 6y, fxy = fyx = ey + 4x3 ,
(3) fxx = 2, fxy = −2, fyy = 6y
(4) zxx = −4 sin(2x − 3y), zxy = 6 sin(2x − 3y), zyy = −9 sin(2x − 3y)
(5) fxx = 0, fxy = 2e2y , fyy = 4xe2y
(6) zxx =
2(1−x2 +y 2 )
(1+x2 +y 2 )2 , zxy
=
−4xy
(1+x2 +y 2 )2 , zyy
=
2(1+x2 −y 2 )
(1+x2 +y 2 )2 ,
(7) fxx = ex sin y + ey cos x, fxy = ex cos y + ey sin x, fyy = −ex sin y − ey cos x
(8) zxx =
−6y
(x+2y)3 , zxy
=
3(x−2y)
(x+2y)3 , zyy
=
12x
(x+2y)3

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