答案是1/e
当x = 1,y = ln(0*1 + e) = lne = 1
所以(0,1)在曲线上.
y = ln(xy+e)
y' = 1/(xy+e) * (y + x*y')
y' = y/(xy+e) + x/(xy+e) * y'
y' * [1 - x/(xy+e)] = y/(xy+e)
y' = y/(xy+e) * (xy+e)/(xy+e-x) = y/(xy+e-x)
在(0,1)的导数是y'(0) = 1/(0*1+e-0) = 1/e