How do you differentiate #f(x)=sec(1/e^(x-4) ) # using the chain rule?

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Answer 1 · 2018-04-20T09:38:17

show the answer below

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]#