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How do you find the derivative of ##sqrt(e^(2x) +e^(-2x))##?
You would use the .
##y=(f(x))^n## ##y'= n(f(x))^(n-1)*f'(x)##
##y= sqrt(e^(2x) + e^(-2x))##
write it in a form that is more understandable
##y = (e^(2x) + e^(-2x))^(1/2)##
lets work out f'(x): remember: ##y = e^(ax)## ##y'(x) = ae^(ax)##
##f(x) = e^(2x) + e^(-2x)## ##f'(x) = 2e^(2x) + (-2)e^(-2x)## ##f'(x) = 2e^(2x) - 2e^(-2x)##
now differentiate: ##y'= (1/2)(e^(2x) + e^(-2x) )^((1/2)-1)* 2e^(2x) - 2e^(-2x)## ##y'= (2e^(2x) - 2e^(-2x))/(2*sqrt(e^(2x) + e^(-2x) ))##
you can simplify it further
##y'= (e^(2x) - e^(-2x))/(sqrt(e^(2x) + e^(-2x))##
:)