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

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x=0
则汪和lny=0
y=1

两边对x求导
[1/(x²+y)]*(x²+y)'=3x²+cosx
(2x+y'镇陵并)/(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/御迹dx(x=0)=1

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