Y=(1-x)/(1+x)的n阶导数怎么求?

y=(1-x)/(1+x)=-1+2/(x+1)=-1+2(x+1)^(-1)
所以y'=-2(x+1)^(-2)
y"=4(x+1)^(-3)
y'''=-12(x+1)^(-4)
所以y(n)=-2*n!*(x+1)^[-(n+1)]
即y(n)=-2*n!/(x+1)^(n+1)