Find y' for y = arcosh(ln x), x > e
Find $y'$ for $y = \operatorname{arcosh}(\ln x)$, $x \gt e$.
Explanation of Method
An inverse hyperbolic cosine of a logarithm.
Step 1: Apply the chain rule
With $u = \ln x$, $u' = \dfrac1x$, and $\bigl(\operatorname{arcosh} u\bigr)' = \dfrac{1}{\sqrt{u^{2}-1}}$: $y' = \frac{1}{\sqrt{(\ln x)^{2}-1}}\cdot\frac{1}{x} = \frac{1}{x\sqrt{\ln^{2}x-1}}$.
Step 2: Verification
\[x = e^{2} \approx 7.38906:\quad \ln x = 2 \;\Longrightarrow\; y' = \frac{1}{e^{2}\sqrt{3}} \approx 0.07814\]
\[\frac{y(e^{2}+10^{-4})-y(e^{2}-10^{-4})}{2\times 10^{-4}} \approx 0.07814 \;\;\checkmark\]
Step 3: Domain
The inverse hyperbolic cosine requires its argument to be at least $1$, i.e. $\ln x \geq 1$, so $x \geq e$. The derivative excludes the endpoint $x = e$, where the radical vanishes and the graph has a vertical tangent.
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{x\sqrt{\ln^{2}x-1}}, \qquad x > e\,}\]
Remark
Remark. Compare the circular case: $\arccos$ needs $|u| \leq 1$ and gives $\sqrt{1-u^{2}}$ in the denominator, while $\operatorname{arcosh}$ needs $u \geq 1$ and gives $\sqrt{u^{2}-1}$. The order of the two terms under the root is the whole difference.