Question

已知y=ln[(1+t)/(1-t)],求y的n阶导数

Answer 1

y=ln[(1+t)/(1-t)]
=ln(1+t)-ln(1-t)
[ln(1+t)]'=1/(t+1)
[ln(1-t)]'=-1/(1-t)
y'=1/(t+1)+1/(1-t)
[1/(t+1)]'=-1/(t+1)^2
[1/(t+1)]''=2/(t+1)^3
[1/(t+1)]^(n)=(-1)^(n)*n!/(t+1)^(n+1)
[1/(1-t)]'=-1/(1-t)^2
[1/(1-t)]''=-2/(1-t)^3
[1/(1-t)]^(n)=-n!/(1-t)^(n+1)
所以
[ln(1+t)/(1-xt]^(n)
=(-1)^(n+1)*(n-1)!/(t+1)^(n)+(n-1)!/(1-t)^(n)
供参考