Question

高等数学:第1题的(1)(2)(3)(4)怎么做，急，需要详细过程

Answer 1

追答

望采纳，谢谢啦

追问

第(1)小题那个:什么意思

追答

是一个点，乘

Answer 2

(1) y'cosy+e^x -y^2-2xyy' = 0, y' = (y^2-e^x)/(cosy-2xy)
(2) (x+2y+z)^2 = 4xyz
得 2(x+2y+z)(1+z'<x>) = 4y(z+xz'<x>)
2(x+2y+z)(2+z'<y>) = 4x(z+yz'<y>)
则 z'<x> = (2yz-x-2y-z)/(x+2y+z-2xy)
z'<y> = 2(xz-x-2y-z)/(x+2y+z-2xy)
(3) x = 1, y = 0 时 z = -1
2yzz'<x> - z^3 - 3xz^2z'<x> = 0
0 -(-1)^3 - 3(-1)^2z'<x> = 0 ,
x = 1, y = 0 时 z'<x> = 1/3
(4) z = e^(2x-3z) + 2y
得 z'<x> = (2-3z'<x>) e^(2x-3z),
z'<y> = -3z'<y> e^(2x-3z)+2
则 z'<x> = 2e^(2x-3z)/[1+3e^(2x-3z)]
z'<y> = 2/[1+3e^(2x-3z)]
3z'<x>+z'<y> = 2[1+3e^(2x-3z)]/[1+3e^(2x-3z)] = 2