高数微分习题

Question

求下列各函数的微分dy
(1)y=3x^2-ln 1/x
(2)y=e^-x cosx

设由下列方程确定y是x的函数，求dy
(1)2x^2 y-xy^2 +y^3=0
麻烦各位给下解题过程，非常感谢~

Answer 1

(1)y = 3x^2 - ln 1/x = 3x^2 + lnx
dy = 6xdx + (1/x)dx
=(6x + 1/x)dx

(2)y = e^(-x)cosx
dy = -e^(-x)cosxdx - e^(-x)sinxdx
= -e^(-x)[cosx + sinx]dx

(3)2x^2 y - xy^2 + y^3=0
4xydx + 2x^2dy - y^2dx - 2xydy + 3y^2dy = 0
dy = [4xy - y^2]dx/[2xy - 3y^2 - 2x^2]
= [(4x - y)y/(2xy - 3y^2 - 2x^2)]dx

Answer 2

(1)dy=6x+1/x
(2)dy=sinxe^(-x)-cosxe^-x
(3)dy=(y^2-4xy)/(2x^2+3y^2-2xy)

Answer 3

如图