#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 5 sub question (ii)

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 the range of $\theta$

\begin{aligned} &x \in\left(\frac{1}{2 \sqrt{2}}, \frac{1}{2}\right) \\\\ &2 x \in\left(\frac{1}{\sqrt{2}}, 1\right) \\\\ &\cos \theta \in\left(\frac{1}{\sqrt{2}}, 1\right) \end{aligned}

\begin{aligned} &\theta \in\left(0, \frac{\pi}{4}\right) \\\\ &2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=\sin ^{-1}(\sin 2 \theta) \end{aligned}

$=2 \theta \quad\left[\begin{array}{l} 2 x=\cos \theta \\ \theta=\cos ^{-1}(2 x) \end{array}\right]$

\begin{aligned} &u=2 \cos ^{-1}(2 x) \\\\ &\frac{d u}{d x}=2\left[\frac{-1}{\sqrt{1-(2 x)^{2}}} \times 2\right] \end{aligned}

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

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

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

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