"求函数 y=ln(1+x/1-x)的n阶导数的一般表达式"这个题该怎么做(我想要看看过程)

y=ln[(1+x)/(1-x)]
=ln(1+x)-ln(1-x)
[ln(1+x)]'=1/(x+1)
[ln(1-x)]'=-1/(1-x)
y'=1/(x+1)+1/(1-x)
[1/(x+1)]'=-1/(x+1)^2
[1/(x+1)]''=2/(x+1)^3
[1/(x+1)]^(n)=(-1)^(n)*n!/(x+1)^(n+1)
[1/(1-x)]'=-1/(1-x)^2
[1/(1-x)]''=-2/(1-x)^3
[1/(1-x)]^(n)=-n!/(1-x)^(n+1)
所以
[ln(1+x)/(1-x)]^(n)
=(-1)^(n+1)*(n-1)!/(x+1)^(n)+(n-1)!/(1-x)^(n)