Question

The derivative of ${\sec }^{-1}\left(\frac{1}{1-2x^2}\right)$w.r.t. sin^-1(3x - 4x³) is

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

Answer 1

The Correct Option is D

Explanation:
\(u=sec^{-1}(\frac{1}{1-2x^{2})}, v=sin^{-1}(3x-4x^3)\)
⇒ \(u=2\theta\), \(v=3\theta\)
Let \(x=sin\theta\)
On differentiating
⇒ \(u=sec^{-1}(sec2\theta)\), \(v=sin^{-1}(sin3\theta)\)
w.r.t. \(\theta\) respectively
\(\frac{du}{d\theta}=\frac{2}{\sqrt{1-x^2}}\), \(\frac{dv}{d\theta}=\frac{3}{\sqrt{1-x^2}}\)
\(\therefore \frac{du}{d\theta}=\frac{\frac{du}{dv}}{d\theta}=\frac{2}{3}\)