Question

1.设y=(x)是由方程xy+lnx=0确定的函数求dy 2.求函数y=x^2/1+x^2的单调区间和极值

Answer 1

1. xy+lnx=0,两边对x求导,y+x*y’+1/x=0,y’=-(y+1/x)/x=-(xy+1)/x^2,则dy=-(xy+1)/x^2*dx

2. y=x^2/(1+x^2) =1-1/(1+x^2),对x求导,y’=2x/(1+x^2)^2, y’’=(2-6x^2)/(1+x^2)^3

令y’=2x/(1+x^2)^2=0,解得：x=0, y’’(0)=2>0,有极小值y=0

单调区间为：x∈(-∞,0)函数递减； x∈(0,+∞)函数递增；极小值点为(0,0)