How do you differentiate #f(x)=sec(1/e^(x-4) ) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer James Apr 20, 2018 show the answer below Explanation: suppose: #u=1/(e^x-4)# #y=sec(u)# #(du)/dx=-e^x/(e^x-4)^2# #(dy)/(du)=sec(u)*tan(u)# #(dy)/dx=(du)/dx*(dy)/(du)# #(dy)/dx=[(-e^x)*(sec(u)*tan(u))]/(e^x-4)^2# #(dy)/dx=[(-e^x)*[sec(1/[e^x-4] )*tan(1/[e^x-4])]/(e^x-4)^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 1357 views around the world You can reuse this answer Creative Commons License