Question

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

• A. $-4$
• B. $4$
• C. $2$
• D. $-2$

Answer 1

Answer: (B)

$4$

Let $u={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

$={\cos }^{-1}(2x^2-1)$ Let $x=\cos \theta$

$={\cos }^{-1}(2{\cos }^2\theta -1)$

$={\cos }^{-1}\left(\cos 2\theta \right)$

$u=2\theta$

$\Rightarrow u=2{\cos }^{-1}x$

$\Rightarrow \frac{du}{dx}=2\times \frac{-1}{\sqrt{1-x^2}}=\frac{-2}{\sqrt{1-x^2}}$

Let $v=\sqrt{1-x^2}$

$\Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1-x^2}}x-2x=\frac{-x}{\sqrt{1-x^2}}$

$\therefore \frac{du/dx}{dv/dx}=\frac{2/\sqrt{1-x^2}}{x/\sqrt{1-x^2}}=\frac{2}{x}$

$\Rightarrow \frac{du}{dv}=\frac{2}{x}$

$\Rightarrow \left(\frac{du}{dv}\right)=\frac{2}{1/2}=2\times 2=4$

$\therefore {\left(\frac{du}{dv}\right)}_{x=1/2}=4$