Find y' for y = arsinh (x-1)/(x+1), x ≠ -1
Find $y'$ for $y = \operatorname{arsinh}\dfrac{x-1}{x+1}$, $x \neq -1$.
Explanation of Method
An inverse hyperbolic sine of a Möbius quotient.
Step 1: Differentiate the inner quotient
With $u = \dfrac{x-1}{x+1}$: $u' = \frac{(x+1)-(x-1)}{(x+1)^{2}} = \frac{2}{(x+1)^{2}}$.
Step 2: Compute $\sqrt{u^{2}+1}$
The cross terms cancel when the two squares are added: \[u^{2}+1 = \frac{(x-1)^{2}+(x+1)^{2}}{(x+1)^{2}} = \frac{2x^{2}+2}{(x+1)^{2}} = \frac{2\bigl(x^{2}+1\bigr)}{(x+1)^{2}},\]
\[ \sqrt{u^{2}+1} = \frac{\sqrt{2\bigl(x^{2}+1\bigr)}}{|x+1|}. \]
The absolute value is essential: $\sqrt{(x+1)^{2}} = |x+1|$, not $x+1$.
Step 3: Combine
Using $\bigl(\operatorname{arsinh} u\bigr)' = \dfrac{u'}{\sqrt{u^{2}+1}}$ and $\dfrac{|x+1|}{(x+1)^{2}} = \dfrac{1}{|x+1|}$: \[y' = \frac{|x+1|}{\sqrt{2\bigl(x^{2}+1\bigr)}}\cdot\frac{2}{(x+1)^{2}} = \frac{2}{|x+1|\sqrt{2\bigl(x^{2}+1\bigr)}} = \frac{\sqrt{2}}{|x+1|\sqrt{x^{2}+1}}.\]
Step 4: Verification
\[x = 1:\quad y = \operatorname{arsinh} 0 = 0,\quad y' = \frac{\sqrt2}{2\sqrt2} = \frac12 \;\;\checkmark\]
\[x = 0:\quad y' = \frac{\sqrt2}{1\cdot 1} \approx 1.41421; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.41421 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{\sqrt{2}}{|x+1|\sqrt{x^{2}+1}}, \qquad x \neq -1\,}\]
Remark
Remark. The derivative is positive on both branches, so the function increases on $(-\infty,-1)$ and on $(-1,\infty)$ separately. Since $\dfrac{x-1}{x+1} \to 1$ as $x \to \pm\infty$, the graph approaches the horizontal asymptote $y = \operatorname{arsinh} 1 = \ln\bigl(1+\sqrt2\bigr)$ in both directions.