y=In √[(2x-1)⼀(x+1)],求导?

2025-06-25 14:22:23
推荐回答(1个)
回答1:

y=ln √[(2x-1)/(x+1)]
=(1/2)(ln(2x-1)-ln(x+1))
y'=(1/2)((2/(2x-1))-(1/(x+1))
=3/(2(2x-1)(x+1))