let
If we can find x in terms of y, then we have found
multlipy both sides by
so
now apply the quadratic forumula to find
we only need the positive root because is always positive
and then take the ln of both sides to arrive at the answer
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
let
If we can find x in terms of y, then we have found
multlipy both sides by
so
now apply the quadratic forumula to find
we only need the positive root because is always positive
and then take the ln of both sides to arrive at the answer