设曲线y=y（x）由方程y-x=e^(xy)确定,求该曲线上在x=0所对应的点出的切线方程

Question

设曲线y=y（x）由方程y-x=e^(xy)确定,求该曲线上在x=0所对应的点出的切线方程

Answer 1

y-x = e^(xy)
y(0) = e^0 =1
y-x = e^(xy)
y' -1 = (xy' +y) e^(xy)
[1-xe^(xy)] y' = 1+ ye^(xy)
y' = [1+ye^(xy)]/[1-xe^(xy)]
y'|(0,1) = 2
曲线上在x=0所对应的点出的切线方程
y -1= 2x