Question

Derivative of ${\sec }^{-1}\left(\frac{1}{1-2x^2}\right)$ w.r.t. ${\sin }^{-1}(3x-4x^3)$ is:

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

Answer 1

Answer: (D)

$u={\sec }^{-1}\left(\frac{1}{1-2x^2}\right),v={\sin }^{-1}(3x-4x^3)$ Let $x=\sin \theta$ $\Rightarrow$ $u{\sec }^{-1}(\sec 2\theta ),v={\sin }^{-1}(\sin 3\theta )$ $\Rightarrow$ $u=2\theta ,v=3\theta$ $\Rightarrow$ $u=2{\sin }^{-1}x,v=3{\sin }^{-1}x$ On differentiating w.r.t. $\theta$ respectively $\frac{du}{d\theta }=\frac{2}{\sqrt{1-x^2}},\frac{dv}{d\theta }=\frac{3}{\sqrt{1-x^2}}$ $\therefore$ $\frac{du}{d\theta }=\frac{\frac{\frac{du}{d\theta }}{dv}}{d\theta }=\frac{2}{3}$