If \(y=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}\), then the ratio \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx} \mathrm{x}^{2}}: \frac{\mathrm{dy}}{\mathrm{dx}}\) is

\(\mathrm{y}=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}\),
Put \(\left\{\begin{array}{l}\mathrm{x}=\sec \theta \\ \mathrm{dx}=\sec \theta \cdot \tan \theta \mathrm{d} \theta\end{array}\right.\)
\(\mathrm{y}=\tan ^{-1} \sqrt{\sec ^{2} \theta-1}=\tan ^{-1}(\tan \theta)=\theta\)
\(=\sec ^{-1} \mathrm{x}\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sec ^{-1} \mathrm{x}\right)=\frac{1}{\mathrm{x} \sqrt{\mathrm{x}^{2}-1}}\)
\(\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}} &=\frac{1}{\mathrm{x}} \cdot \frac{-1}{2} \frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}(2 \mathrm{x})-\frac{1}{\mathrm{x}^{2}} \cdot \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \\ \quad=-\frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{1 / 2}} \\=-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}}\left(\mathrm{x}^{2}+\mathrm{x}^{2}-1\right) \\ \quad=-\frac{\left(2 \mathrm{x}^{2}-1\right)}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}} \end{aligned}\)
\[
\begin{aligned}
&\text { Now, } \frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}: \frac{1}{x\left(x^{2}-1\right)^{1 / 2}} \\
&\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=\left(1-2 x^{2}\right): x\left(x^{2}-1\right) \\
&\text { or } \quad\left(\frac{d^{2} y}{\frac{d x^{2}}{d y}}\right)=\frac{\left(1-2 x^{2}\right)}{x\left(x^{2}-1\right)}
\end{aligned}
\]