How do you differentiate # y = 2/[(e^(x) + e^(-x)]#? Calculus Basic Differentiation Rules Chain Rule 1 Answer sjc Oct 17, 2017 #(dy)/(dx)=-(2(e^x-e^(-x)))/((e^x+e^(-x))^2# Explanation: we will use teh quotient rule #y=u/v=>y'=(vu'-uv')/v^2# #y=2/(e^x+e^(-x)# #u=2=>u'=0# #v=e^x+e^(-x)=>v'=e^x-e^(-x)# #;.y'=(0-2(e^x-e^(-x)))/(e^x+e^(-x))^2# #(dy)/(dx)=-(2(e^x-e^(-x)))/((e^x+e^(-x))^2# 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 1339 views around the world You can reuse this answer Creative Commons License