Question:

At what point of the curve y=cosh (x) does the tangent have slope 1?

It seems so simple but I'm having a hard time working through this one. I know I need to find X when sinh (x)=1. I looked at the answer however and it is;

(ln(1+sqr(2)), sqr(2))

The X term in this point is similar to the arcsinh (x)= ln(x+sqr(x^2 + 1)). I don't understand how it got there? Any help walking through this one??

Thanks all,

cheers