如果f'(x)=sin x^2 ,y=f(2x/x-1),求dy/dx

网友回答

dy/dx = y'
= f'[ 2x / (x-1) ] * [ [ 2x / (x-1) ] '
= sin[ 2x / (x-1) ]² * -2 / (x-1)²
= -2 sin[ 2x / (x-1) ]² / (x-1)²