设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx

复合函数求导
设 y=f(t),t(x)=(2x-1)/(x+1)
则
dy/dt = lnt^(1/3)=ln{[(2x-1)/(x+1)]^(1/3)},
dt/dx=[(2x-1)/(x+1)]'=3/(x+1)^2
【具体过程】
dy/dx
=(dy/dt)*(dt/dx)
=f'[(2x-1)/(x+1)]*[(2x-1)/(x+1)]'
=ln{[(2x-1)/(x+1)]^(1/3)}*[3/(x+1)^2]
=(1/3)*ln[(2x-1)/(x+1)]*[3/(x+1)^2]
=[ln(2x-1)-ln(x+1)]/(x+1)^2

令u=（2x-1)/(x+1）
所以f'(u)=lnu^(1/3)
则dy/dx=df(u)/dx=f'(u)u'(x)=1/3*ln(（2x-1)/(x+1）)*u'(x)
下面的过程应该很清楚了吧