Question

设y=(x)是由方程x^3+y^3-3axy=0所确定,求dy

Answer 1

x=0

则lny=0

y=1

两边对x求导

[1/(x²+y)]*(x²+y)'=3x²+cosx

(2x+y')/(x²+y)=3x²+cosx

y'=(x²+y)(3x²+cosx)-2x

即dy/dx=(x²+y)(3x²+cosx)-2x

x=0,y=1

所以dy/dx(x=0)=1