Question

The derivative of $se{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=0$, is

• A. $2$
• B. $-2$
• C. $1$
• D. non existent

Answer 1

The correct option is D non existent

Let $u=se{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $v=\sqrt{1-x^2}$ at $x=0$
We have to find $\frac{du}{dv}$
$\frac{du}{dx}=\frac{d}{dx}\left[{\sec }^{-1}\left(\frac{1}{2x^2-1}\right)\right]$
put $x=\cos \theta$
we get,
$\frac{du}{dx}=\frac{d}{dx}\left[{\sec }^{-1}\left(\frac{1}{2{\cos }^2\theta -1}\right)\right]$
$=\frac{d}{dx}\left[{\sec }^{-1}\left(\frac{1}{\cos 2\theta }\right)\right]$
$=\frac{d}{dx}\left[{\sec }^{-1}\left(\sec 2\theta \right)\right]$
$=\frac{d}{dx}\left(2\theta \right)$
$=\frac{d}{dx}\left(2{\cos }^{-1}x\right)$
$=\frac{-2}{\sqrt{1-x^2}}$..........(1)

$\frac{dv}{dx}=\frac{d}{dx}\left(\sqrt{1-x^2}\right)$

$=\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}(1-x^2)$
$=\frac{-x}{\sqrt{1-x^2}}$ ............(2)
From 1 and 2
$\frac{du}{dv}=\frac{-2}{\sqrt{1-x^2}}\times \frac{\sqrt{1-x^2}}{-x}$
$=\frac{2}{x}$ at $x=0$
$\frac{du}{dv}=\frac{2}{0}=$ not existing