How do you differentiate #f(x)=sqrtcsc(e^(4x))# using the chain rule.?

1 Answer
Mar 12, 2018

Answer:

#(dy)/(dx)=-(cot(e^(4x))sqrt(csc(e^(4x))))/2# #4e^(4x)#

Explanation:

let #u=e^(4x)# so that #(du)/dx=4e^(4x)#

and let #y=f(x)# so that #y=(csc(u))^(1/2)# so#(dy/(du))=-(cot(u)sqrt(csc(u)))/2#

then by the chain rule

#(dy)/(dx)=(dy)/(du).(du)/(dx)#

so
#(dy)/(dx)=-(cot(e^(4x))sqrt(csc(e^(4x))))/2# #4e^(4x)#