How do you find the derivative of #sqrt(e^(2x) +e^(-2x))#?

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Answer 1 · 2016-07-22T19:37:07

#(dy)/(dx) = (e^(2x) - e^(-2x))/(sqrt(e^(2x) + e^(-2x))#

#y = (e^(2x) + e^(-2x))^(1/2)#

We need to use the chain rule as we have #y(u(x))#.

In this case #(dy)/(dx) = (dy)/(du)(du)/(dx)#

#u = e^(2x) + e^(-2x) implies (du)/(dx) = 2e^(2x) - 2e^(-2x)#

#(dy)/(du) = 1/2(e^(2x) + e^(-2x))^(-1/2)# using the power rule

Hence:

#(dy)/(dx) = 1/2(e^(2x) + e^(-2x))^(-1/2)*2e^(2x) - 2e^(-2x)#

#(dy)/(dx) = (e^(2x) - e^(-2x))/(sqrt(e^(2x) + e^(-2x))#