Question

设函数y=y(x)由方程y+x=e的xy次方确定,求y'(0)

Answer 1

y+x=e^xy x=0 y=1

y'+1=(y+xy')e^xy

1-ye^xy=y'(xe^xy-1)

y'=(1-ye^xy)/(xe^xy-1)

y'(0)=(1-1)/(-1)=0