Question

Find $\frac{dy}{dx}$ if $y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$.

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{-2}{\sqrt{1-x^2}}$
• C. $\frac{1}{1+x^2}$
• D. None of these

Answer 1

The correct option is B

$\frac{-2}{\sqrt{1-x^2}}$

Step-1: Find the required derivative:

Given: $y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$.

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

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

Now, Denominator

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

$=\cos \left(2\theta \right)$

We can write,

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

Step-2: Differentiating with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{d2{\cos }^{-1}\left(x\right)}{dx}\left[\because \frac{d}{dx}{\cos }^{-1}\left(x\right)=-\frac{1}{\sqrt{1-x^2}}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}} \end{gathered}$$

Hence, the correct answer is $\frac{-2}{\sqrt{1-x^2}}$.