Question

The derivative of sec−1⁡(11−2x2)w.r.t. sin-1(3x - 4x3) is

• A. 14
• B. 32
• C. 1
• D. 23

Answer 1

Answer: (D)

23

Answer 2

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
$\frac{du}{d\theta}=\frac{2}{\sqrt{1-x^2}}$, $\frac{dv}{d\theta}=\frac{3}{\sqrt{1-x^2}}$
$\therefore \frac{du}{d\theta}=\frac{\frac{du}{dv}}{d\theta}=\frac{2}{3}$