设函数y=f(x)由方程ln(x^2+y)=x^3+sinx确定,求dy/dx(x=0)

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