求解一道导数题(微分),数学大神们看一下,谢谢。

2025-05-08 02:56:33
推荐回答(1个)
回答1:

f=(1/(x^2-1))'=1/2(1/(x-1)-1/(x+1))'=1/2[1/(x+1)^2-1/(x-1)^2]
f'=-[1/(x+1)^3-1/(x-1)^3]
f''=3[1/(x+1)^4-1/(x-1)^4]