设隐函数y=y(x)由方程xy+Iny=1确定,令g(x)=y(Inx)(e^y(x)),求g’(1)

xy + lny = 1
当 x = 1,y + lny = 1, 无法求解,估计得到 y =1;
当 x = 0,y = e
y + xy' + (1/y)y'= 0
y²+ xyy'+ y'= 0
y' = -y²/(xy + 1)
当 x = 0 和 y = e,
得 y'= -e²;
当 x = 1 和 y = 1,
得 y'= -1/(1+1) = -1/2
g(x) = y(lnx)(e^y)
g'(x)=[y'(lnx)](1/x)e^y+y(lnx)(e^y)y'
g'|(x=1)=
= [y'(ln1)](1/1)e^1 + y(ln1)(e^1)(-1/2)
= [y'(0)]e - ½e×y(0)
= -e³-½e²

y'=-y²/(xy+1)
所以x=1时
y=1,y'=-1/2
g'(x)=e^y[y'(lnx)(1+y)+y/x]
所以
g'(1)=e
答案错了吧……

楼上正解，一楼算g(x)的导数错了