Question

Differential coefficient of ${\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{\left(1-x^2\right)}$ at $x=\frac{1}{2}$ is

• A. $2$
• B. $4$
• C. $6$
• D. $1$

Answer 1

The correct option is B

$4$

Explanation for correct option:

Step-1: Differentiate ${\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $x$.

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

Put, $x=\cos \left(\theta \right)$

$$\begin{gathered} \Rightarrow u={\sec }^{-1}\left(\frac{1}{2{\left(\cos \left(\theta \right)\right)}^2-1}\right) \\ \Rightarrow u={\sec }^{-1}\left(\frac{1}{\cos \left(2\theta \right)}\right)\left(\because 2{\left(\cos \left(\theta \right)\right)}^2-1=\cos \left(2\theta \right)\right) \\ \Rightarrow u={\sec }^{-1}\left(\sec \left(2\theta \right)\right) \\ \Rightarrow u=2\theta \\ \Rightarrow u=2{\cos }^{-1}\left(x\right) \end{gathered}$$

Differentiate with respect to $x$.

$\Rightarrow \frac{du}{dx}=-\frac{2}{\sqrt{\left(1-x^2\right)}}........\left(i\right)$

Step-2: Differentiate $\sqrt{\left(1-x^2\right)}$ with respect to $x$.

Given, $v=\sqrt{\left(1-x^2\right)}$

Differentiate with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{\left(1-x^2\right)}}\times \left(0-2x\right) \\ \Rightarrow \frac{dv}{dx}=-\frac{x}{\sqrt{\left(1-x^2\right)}}.......\left(ii\right) \end{gathered}$$

Step-3: Divide equation $\left(i\right)$ by equation $\left(ii\right)$.

$$\begin{gathered} \Rightarrow \frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)}=\frac{\left(-{\displaystyle \frac{2}{\sqrt{\left(1-x^2\right)}}}\right)}{\left(-{\displaystyle \frac{x}{\sqrt{\left(1-x^2\right)}}}\right)} \\ \Rightarrow \frac{du}{dv}=\frac{2}{x} \end{gathered}$$

Step-4: Put $x=\frac{1}{2}$

$$\begin{gathered} \Rightarrow \frac{du}{dv}=\frac{2}{\left({\displaystyle \frac{1}{2}}\right)} \\ \Rightarrow \frac{du}{dv}=4 \end{gathered}$$

Hence, correct answer is option $\left(B\right)$