Find y' for y = (arcosh x)/(ln x), x > 1
Find $y'$ for $y = \dfrac{\operatorname{arcosh} x}{\ln x}$, $x \gt 1$.
Explanation of Method
A quotient of an inverse hyperbolic function and a logarithm.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = \operatorname{arcosh} x,\quad u' = \frac{1}{\sqrt{x^{2}-1}}, \qquad v = \ln x,\quad v' = \frac{1}{x}. \]
The condition $x \gt 1$ is exactly what makes both functions defined and $\ln x \neq 0$.
Step 2: Substitute
\[ y' = \frac{\dfrac{\ln x}{\sqrt{x^{2}-1}} - \dfrac{\operatorname{arcosh} x}{x}}{\bigl(\ln x\bigr)^{2}}. \]
Step 3: Clear the inner fractions
Multiplying numerator and denominator by $x\sqrt{x^{2}-1}$ gives a single fraction: \[y' = \frac{x\ln x - \sqrt{x^{2}-1}\,\operatorname{arcosh} x}{x\sqrt{x^{2}-1}\,\bigl(\ln x\bigr)^{2}}.\]
Step 4: Verification
\[x = 2:\quad \operatorname{arcosh} 2 \approx 1.31696,\ \ln 2 \approx 0.69315,\ \sqrt{3} \approx 1.73205\]
\[y' \approx \frac{0.69315/1.73205 - 1.31696/2}{0.69315^{2}} = \frac{0.40019 - 0.65848}{0.48045} \approx -0.53761\]
\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx -0.53761 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \frac{\dfrac{\ln x}{\sqrt{x^{2}-1}} - \dfrac{\operatorname{arcosh} x}{x}}{\bigl(\ln x\bigr)^{2}} = \frac{x\ln x - \sqrt{x^{2}-1}\,\operatorname{arcosh} x}{x\sqrt{x^{2}-1}\,\bigl(\ln x\bigr)^{2}}\,}\]
Remark
Remark. In logarithmic form $\operatorname{arcosh} x = \ln\bigl(x+\sqrt{x^{2}-1}\bigr)$, so $y$ is a ratio of two logarithms. As $x \to 1^{+}$ the numerator $\operatorname{arcosh} x$ vanishes like $\sqrt{2(x-1)}$ while $\ln x$ vanishes like $x-1$, so $y \to +\infty$ — the square root wins.