Question

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

Short answer

The point is \(\left( {\ln \left( {1 + \sqrt 2 } \right),\sqrt 2 } \right)\).

Step-by-step solution

Write the formula of derivatives of hyperbolic functions

\(\frac{d}{{dx}}\left( {\cosh x} \right) = \sinh x\)

Find the point of the curve \(y = coshx\)

Consider the curve \(y = \cosh x\).

The tangent to \(y = \cosh x\) has slope \(1\) .

Therefore,

\(y' = \sinh x\)

Now,

\(\begin{aligned}\sinh x & = 1\\x & = {\sinh ^{ - 1}}1\\ & = \ln \left( {1 + \sqrt 2 } \right)\end{aligned}\)

Since \(\sinh x = 1\) and \(y = \cosh x\).

Then we have,

\(\begin{aligned}y & = \cosh x\\ & = \sqrt {1 + {{\sinh }^2}x} \end{aligned}\)

Since \(\cosh x = \sqrt 2 \). This implies that, \(y = \sqrt 2 \).

Thus, the point is \(\left( {\ln \left( {1 + \sqrt 2 } \right),\sqrt 2 } \right)\).