Examine the sign of the second derivative of y = arcsin

Examine the sign of the second derivative of $y = \arcsin\dfrac{1}{x-1}$.

Step 1: Domain

The arcsine requires $\left|\dfrac{1}{x-1}\right|\le1$, i.e. $|x-1|\ge1$: $\text{domain } = \bigl(-\infty,\,0\bigr]\cup\bigl[2,\,+\infty\bigr)$. two branches that never meet. Write $t = x-1$, so $|t|\ge1$; differentiation is carried out on the open sets $|t|\gt 1$.

Step 2: The first derivative

With $u = \dfrac1t$, $u' = -\dfrac{1}{t^{2}}$ and $\sqrt{1-u^{2}} = \dfrac{\sqrt{t^{2}-1}}{|t|}$, \[y' = \frac{u'}{\sqrt{1-u^{2}}} = -\frac{1}{t^{2}}\cdot\frac{|t|}{\sqrt{t^{2}-1}} = \frac{-1}{|t|\sqrt{t^{2}-1}}. \tag{1}\]

So $y'\lt 0$ on both branches: the function decreases on each.

Step 3: The second derivative on the right branch $\bigl(t\gt 1\bigr)$

There $|t| = t$, so $y' = -t^{-1}\bigl(t^{2}-1\bigr)^{-1/2}$ and

\[ y'' = \frac{1}{t^{2}\sqrt{t^{2}-1}}+\frac{1}{\bigl(t^{2}-1\bigr)^{3/2}} = \frac{2t^{2}-1}{t^{2}\bigl(t^{2}-1\bigr)^{3/2}} \gt 0, \tag{2} \] since $2t^{2}-1\gt 1$ when $t^{2}\gt 1$.

Step 4: The left branch $\bigl(t\lt -1\bigr)$ is not the same

There $|t| = -t$, so (1) reads $y' = \dfrac{1}{t\sqrt{t^{2}-1}}$ — the opposite sign in front of the same expression. Differentiating, \[y'' = -\frac{1}{t^{2}\sqrt{t^{2}-1}}-\frac{1}{\bigl(t^{2}-1\bigr)^{3/2}} = -\frac{2t^{2}-1}{t^{2}\bigl(t^{2}-1\bigr)^{3/2}} \lt 0. \tag{3}\]

Both cases are covered by one formula if the sign of $t$ is kept: \[y'' = \frac{2\bigl(x-1\bigr)^{2}-1}{\bigl(x-1\bigr)\bigl|x-1\bigr|\left[\bigl(x-1\bigr)^{2}-1\right]^{3/2}}. \tag{4}\]

Step 5: Conclusion

The numerator $2(x-1)^{2}-1\ge1$ on the domain, so the sign of $y''$ is the sign of $x-1$: \[y''\gt 0 \text{ on } \bigl(2,+\infty\bigr) \ \text{(concave up)}; \qquad y''\lt 0 \text{ on } \bigl(-\infty,0\bigr) \ \text{(concave down)}.\]

There is no inflection point: the two signs occur on different branches, separated by the gap $\bigl(0,2\bigr)$ where the function is not defined, so the concavity never actually reverses along the curve.

Step 6: Verification

\[y''(3) = \frac{7}{4\cdot3^{3/2}} = \frac{7\sqrt3}{36} = 0.3368\gt 0; \qquad y''(-1) = -\frac{7\sqrt3}{36} = -0.3368\lt 0 \;\;\checkmark\]

\[\text{Second differences: } y(4.5)-2y(4)+y(3.5) = +0.0216; \qquad y(-0.5)-2y(-1)+y(-1.5) = -0.0940 \;\;\checkmark\]

\[y(-2) = -0.3398369,\ y(-1) = -0.5235988,\ y(0) = -\tfrac{\pi}{2}:\ \text{falling and bending downward} \;\;\checkmark\]