设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx
问题描述：
设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx
答
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)/...