Choose the correct alternative in the following:

Solution:

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

$\frac{\mathrm{du}}{\mathrm{dv}}=?$

Considering $u=\cos ^{-1}\left(2 x^{2}-1\right)$

Put $x=\cos \theta \Rightarrow \theta=\cos ^{-1} x \cdots$ (1)

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

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

$u=2 \theta$

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

Differentiating w.r.t $x$ we get,

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=-\frac{2}{\sqrt{1-\mathrm{x}^{2}}}$

Considering $v=\cos ^{-1} x$

Differentiating w.r.t $\mathrm{x}$ we get,

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

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dv}}$

$\Rightarrow \frac{d u}{d v}=-\frac{2}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right)$

$\Rightarrow \frac{d u}{d v}=2$