y=In √[(2x-1)/(x+1)],求导?

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y=In√[(2x-1)/(x+1)],求导?y=In√[(2x-1)/(x+1)],求导?y=In√[(2x-1)/(x+1)],求导?y=ln√[(2x-1)/(x+1)]=(1/2)(ln(2x

y=In √[(2x-1)/(x+1)],求导?
y=In √[(2x-1)/(x+1)],求导?

y=In √[(2x-1)/(x+1)],求导?
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))