Question

1、求由方程2y-x=(x-y)ln(x-y)所确定的函数y=y(x)的微分dy

Answer 1

第一题,这是个隐函数,两边对x求导得：

2y'-1=(1-y')*ln(x-y)+(x-y)*(1-y')/(x-y)=(1-y')*ln(x-y)+(1-y')

所以

[3+ln(x-y)]y'=ln(x-y)+2

y'=[ln(x-y)+2]/[ln(x-y)+3]

所以

dy=[ln(x-y)+2]dx/[ln(x-y)+3]

第二题,

令g(x)=f(x)-x,则原命题等价于g(x)=0在(0,1)上至多有一个实根.

所以g'(x)=f'(x)-1