Examine the sign of the second derivative of y = ln

Examine the sign of the second derivative of $y = \ln\dfrac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+x}$.

Step 1: Simplify first

Since $\sqrt{x^{2}+1}\gt |x|$, the argument is positive for every $x$. Multiplying numerator and denominator by $\sqrt{x^{2}+1}+x$: \[\frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+x} = \frac{\bigl(x^{2}+1\bigr)-x^{2}}{\left(\sqrt{x^{2}+1}+x\right)^{2}} = \frac{1}{\left(\sqrt{x^{2}+1}+x\right)^{2}},\]

\[ y = -2\ln\left(x+\sqrt{x^{2}+1}\right) = -2\operatorname{arsinh}x. \tag{1} \]

Step 2: The first derivative

\[ y' = -2\cdot\frac{1+\dfrac{x}{\sqrt{x^{2}+1}}}{x+\sqrt{x^{2}+1}} = -2\cdot\frac{1}{\sqrt{x^{2}+1}} = \frac{-2}{\sqrt{x^{2}+1}}, \] the numerator and denominator sharing the factor $x+\sqrt{x^{2}+1}$. In particular $y'\lt 0$ everywhere: the function is strictly decreasing.

Step 3: The second derivative

\[ y' = -2\bigl(x^{2}+1\bigr)^{-1/2} \;\Longrightarrow\; y'' = \frac{2x}{\bigl(x^{2}+1\bigr)^{3/2}}. \tag{2} \]

Step 4: Sign

The denominator is positive, so $y''$ has the sign of $x$: $x\lt 0:\ y''\lt 0 \ \text{(concave down)}; \qquad x\gt 0:\ y''\gt 0 \ \text{(concave up)}$.

The single sign change at the origin gives one inflection point, and $y(0) = \ln1 = 0$.

Step 5: Verification

\[y(1) = -2\ln\bigl(1+\sqrt2\bigr) = -1.7627472; \qquad y(-1) = +1.7627472 \;\;\checkmark\ \text{(odd function)}\]

\[y'(0) = -2:\ \text{the inflection has an oblique tangent of slope } -2 \;\;\checkmark\]

\[y''(1) = \frac{2}{2\sqrt2} = 0.7071\gt 0; \qquad y''(-1) = -0.7071\lt 0 \;\;\checkmark\]