Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 5 sub question (i) maths textbook solution

Answer: $\frac{-1}{x}$

Hint:   $\text { Let } u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right), v=\sqrt{1-4 x^{2}}$
Given:  $\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \text { w.r.t } \sqrt{1-4 x^{2}}$

Explanation:

$\text { Let } 2 x=\cos \theta$

\begin{aligned} &u=\sin ^{-1}\left(4 x \sqrt{1-4 x^{2}}\right) \\\\ &u=\sin ^{-1}\left(2 \cos \theta \sqrt{1-(\cos \theta)^{2}}\right) \end{aligned}

\begin{aligned} &=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\\\ &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}

Now we try to find range of $\theta$

$x \in\left(\frac{-1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right)$

$2 x \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

$\cos \theta \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\; \; \; \; \; \; \; \; \; \left[\begin{array}{l} \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} \\ \& \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \end{array}\right]$

\begin{aligned} &\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) \\\\ &2 \theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \end{aligned}

\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta) \\\\ &=\pi-2 \theta \\\\ &=\pi-2 \cos ^{-1}(2 x) \end{aligned}                    $\left[\begin{array}{l} 2 x=\cos \theta \\ \theta=\cos ^{-1}(2 x) \end{array}\right]$

$\frac{d u}{d x}=0-2\left[\frac{-1}{\sqrt{1-(2 x)^{2}}} \times 2\right]$

$=\frac{4}{\sqrt{1-x^{2}}}$

\begin{aligned} &v=\sqrt{1-4 x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-4 x^{2}}}(-8 x) \\\\ &=\frac{-4 x}{\sqrt{1-x^{2}}} \end{aligned}

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{4}{\sqrt{1-x^{2}}}}{\frac{-4 x}{\sqrt{1-x^{2}}}}=\frac{-1}{x}$