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))##

🙂