Question

If $y =tan ^{-1} \sqrt {x^2-1}$ then the ratio $\frac {d^2y}{dx^2}: \frac {dy}{dx}$=_________

• A. $\frac {1+2x^2}{x(x^2+1)}$
• B. $\frac {x(x^n+1)}{(1-2x^2)}$
• C. $\frac {x(x^2 -1)}{(1+2x^2)}$
• D. $\frac {1- 2x^2 }{x(x^2-1)}$

Answer 1

The Correct Option is D

$y=\tan ^{-1} \sqrt{x^{2}-1}$ Put $\begin{cases} x=\sec \theta \\ d x=\sec \theta \cdot \tan \theta d \theta \end{cases}$ $y=\tan ^{-1} \sqrt{\sec ^{2} \theta-1}=\tan ^{-1}(\tan \theta)=\theta$ $=\sec ^{-1} x$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^{2}-1}}$ $\frac{d^{2} y}{d x^{2}} =\frac{1}{x} \cdot \frac{-1}{2} \frac{1}{\left(x^{2}-1\right)^{3 / 2}}(2 x)-\frac{1}{x^{2}} \cdot \frac{1}{\sqrt{x^{2}-1}}$ $=-\frac{1}{\left(x^{2}-1\right)^{3 / 2}}-\frac{1}{x^{2}\left(x^{2}-1\right)^{1 / 2}}$ $=-\frac{1}{x^{2}\left(x^{2}-1\right)^{3 / 2}}\left(x^{2}+x^{2}-1\right)$ $=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}$ Now, $\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}$ $=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}: \frac{1}{x\left(x^{2}-1\right)^{1 / 2}}$ $\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=\left(1-2 x^{2}\right): x\left(x^{2}-1\right)$ or $\left(\frac{\frac{d^{2} y}{d x^{2}}}{\frac{d y}{d x}}\right)$ $=\frac{\left(1-2 x^{2}\right)}{x\left(x^{2}-1\right)}$