How do you find the derivative of #y = [e^(-1) + e^(t)]^3#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Shura May 15, 2015 The answer is : #3(e^(-1) + e^t)^2*e^t # #dot y = dot [(e^(-1) + e^t)^3] = 3(e^(-1) + e^t)^2*dot ((e^(-1)+e^t))# #= 3(e^(-1) + e^t)^2*e^t # 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 1705 views around the world You can reuse this answer Creative Commons License