differentiate y = ln(e^x + e^-x)

Answer 1

substitute u=e^x+e^-x

d(log(e^x+e^-x))/dx
=d(log(u))/dx. du/dx
= 1/u * (e^x-e^-x)
= (e^x-e^-x)/(e^x+e^-x)