The derivative of cos^-1(2x^2 - 1) w.r.t. cos^-1x is

The correct answer is Option (1) → $2$ ##

Let $u = \cos^{-1}(2x^2 - 1)$ and $v = \cos^{-1}x$

Put $x = \cos \theta$

$u = \cos^{-1}(2\cos^2 \theta - 1)$

$u = \cos^{-1} \cos 2\theta = 2\theta \quad [∵2\cos^2 \theta - 1 = \cos 2\theta]$

$u = 2\cos^{-1}x$

On differentiating w.r.t. $x$, we get

$\frac{du}{dx} = -\frac{2}{\sqrt{1-x^2}} \quad \left[ ∵\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1-x^2}} \right]$

and $v = \cos^{-1}x$

$\frac{dv}{dx} = -\frac{1}{\sqrt{1-x^2}}$

Now, $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{-2}{\sqrt{1-x^2}}}{\frac{-1}{\sqrt{1-x^2}}}$

$∴\frac{du}{dv} = 2$