Answer
The tangent has a slope of 1 at the point $(ln(1+\sqrt{2}),\sqrt{2})$
Work Step by Step
$y = cosh~x$
$y' = sinh~x = 1$
$\frac{e^x-e^{-x}}{2} = 1$
$e^x-e^{-x} = 2$
$\frac{e^{2x}-1}{e^x} = 2$
$e^{2x}-1 = 2e^x$
$(e^x)^2-2e^x-1 = 0$
We can use the quadratic formula:
$e^x = \frac{2\pm \sqrt{(-2)^2-(4)(1)(-1)}}{2(1)}$
$e^x = \frac{2\pm \sqrt{8}}{2}$
$e^x = 1 \pm \sqrt{2}$
Since $e^x \gt 0$ for all $x$, the solution is $e^x = 1+\sqrt{2}$
Then:
$x = ln(1+\sqrt{2})$
We can find $y$:
$y = cosh~x$
$y = \frac{e^x+e^{-x}}{2}$
$y = \frac{e^{ln(1+\sqrt{2})}+e^{- ln(1+\sqrt{2})}}{2}$
$y = \frac{1+\sqrt{2}+\frac{1}{1+\sqrt{2}}}{2}$
$y = \frac{\frac{3+2\sqrt{2}}{1+\sqrt{2}}+\frac{1}{1+\sqrt{2}}}{2}$
$y = \frac{\frac{4+2\sqrt{2}}{1+\sqrt{2}}}{2}$
$y = \frac{2+\sqrt{2}}{1+\sqrt{2}}\cdot \frac{1-\sqrt{2}}{1-\sqrt{2}}$
$y = \frac{-\sqrt{2}}{-1}$
$y = \sqrt{2}$
The tangent has a slope of 1 at the point $(ln(1+\sqrt{2}),\sqrt{2})$
