How do you simplify cosh(ln x^2)?

We haven't covered this in class yet, but I would like to know how to simplify this.

1 Answer
Feb 25, 2018

#cosh(lnx^2)=(x^4+1)/(2x^2)#

Explanation:

As #coshx=(e^x+e^(-x))/2#, we can write #cosh(lnx^2)#

as #cosh(lnx^2)=(e^(lnx^2)+e^(-lnx^2))/2#

= #(e^(lnx^2)+1/e^(lnx^2))/2#

but as #e^(lna)=a#, we can write it #e^(lnx^2)# as #x^2#

#cosh(lnx^2)=(x^2+1/x^2)/2=(x^4+1)/(2x^2)#