Question

设y=y(x)是由方程y=tan(x+y)所确定的隐函数,求微分dy

Answer 1

两边对x求导：

y'=(1+y')[sec(x+y)]^2

得y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=1/{[cos(x+y)]^2-1}

因此dy=dx/{[cos(x+y)]^2-1}