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))##
🙂
