Find y' for y = ln cosh x

Find $y'$ for $y = \ln\cosh x$.

Explanation of Method

The logarithmic derivative of the hyperbolic cosine.

Step 1: Apply the chain rule

With $u = \cosh x$ and $u' = \sinh x$: \[y' = \frac{u'}{u} = \frac{\sinh x}{\cosh x} = \tanh x.\]

Step 2: Verification

\[x = 0:\quad y = \ln 1 = 0,\quad y' = \tanh 0 = 0, \text{ the minimum of } y \;\;\checkmark\]

\[x = 1:\quad y' = \tanh 1 \approx 0.76159; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.76159 \;\;\checkmark\]

Result

Final answer:

\[\boxed{\,y' = \tanh x\,}\]

Remark

Remark. Since $\cosh x \geq 1 \gt 0$, the function is defined on all of $\mathbb{R}$ — unlike $\ln\cos x$ (Problem 120), which lives only on intervals where the cosine is positive. The absence of a minus sign is the usual hyperbolic-versus-circular difference: $(\cosh x)' = +\sinh x$.