Question

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

• 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

Answer: (D)

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