设函数y=f(x)由方程xy+2lnx=y4所确定,则曲线y=f(x)在点(1,1)处的切线方程是_.

问题描述:

设函数y=f(x)由方程xy+2lnx=y4所确定,则曲线y=f(x)在点(1,1)处的切线方程是______.

等式xy+2lnx=y4两边直接对x求导,得
y+xy′+

2
x
=4y3y′
将x=1,y=1代入上式,有 y'(1)=1 故过点(1,1)处的切线方程为
y-1=1•(x-1),即x-y=0
故答案为:x-y=0