y=arctanu复合函数求导，
arctan'[(1+x)/(1-x)]=[(1+x)/(1-x)]'/{1+[(1+x)/(1-x)]²}=[(1-x+1+x)/(1-x)²]/{1+[(1+x)/(1-x)]²}=2/[(1-x)²+(1+x)²]=2/(2+2x²)=1/(1+x²)
