Question

Derivative of ${\sec }^{-1}\left(\frac{1}{1-2x^2}\right)$ with respect to ${\sin }^{-1}\left(3x-4x^3\right)$ is

• A. $\frac{1}{4}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $1$

Answer 1

The correct option is C

$\frac{2}{3}$

Explanation for correct option:

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

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

Put, $$\begin{gathered} x=\sin \left(\theta \right) \\ \theta ={\sin }^{-1}\left(x\right) \end{gathered}$$

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

Differentiate with respect to $x$

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

Step-2: Differentiate ${\sin }^{-1}\left(3x-4x^3\right)$ with respect to $x$.

Let, $v={\sin }^{-1}\left(3x-4x^3\right)$

Put, $$\begin{gathered} x=\sin \left(\theta \right) \\ \theta ={\sin }^{-1}\left(x\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow v={\sin }^{-1}\left(3\sin \left(\theta \right)-4{\left(\sin \left(\theta \right)\right)}^3\right) \\ \Rightarrow v={\sin }^{-1}\left(\sin \left(3\theta \right)\right)\left(\because 3\sin \left(\theta \right)-4{\left(\sin \left(\theta \right)\right)}^3=\sin \left(3\theta \right)\right) \\ \Rightarrow v=3\theta \\ \Rightarrow v=3{\sin }^{-1}\left(x\right) \end{gathered}$$

Differentiate with respect to $x$

$\Rightarrow \frac{dv}{dx}=\frac{3}{\sqrt{1-x^2}}...............\left(ii\right)$

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

$$\begin{gathered} \Rightarrow \frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)}=\frac{{\displaystyle \frac{2}{\sqrt{1-x^2}}}}{{\displaystyle \frac{3}{\sqrt{1-x^2}}}} \\ \Rightarrow \frac{du}{dv}=\frac{2}{3} \end{gathered}$$

Derivative of ${\sec }^{-1}\left(\frac{1}{1-2x^2}\right)$ with respect to ${\sin }^{-1}\left(3x-4x^3\right)$ is $\frac{2}{3}$.

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