hyperbolic help

[bard]

at what point on the curve y=\cosh x does the tangent have slope 1


I have no idea how to approach this problem

my work

1=\sinh x\frac{dy}{dx}

\frac{1}{sinh x}=\frac{dy}{dx}

[NateTG]

It would probably help to you use the definitions.

For example:
\sinh{x}= \frac{e^{x}-e^{-x}}{2}

[bard]

yah I found the answer as x=ln[1+sqt2]----thanks for your help