How do you differentiate #y=((x+1)/(x-1))^2#? Calculus Basic Differentiation Rules Chain Rule 1 Answer GiĆ³ May 16, 2015 I would use the Chain and Quotient Rule: #y'=2((x+1)/(x-1))((x-1)-(x+1))/(x-1)^2=# #=2((x+1)/(x-1))(-2)/(x-1)^2=# #=-4(x+1)/(x-1)^3# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 2827 views around the world You can reuse this answer Creative Commons License