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Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 5 sub question (iii)

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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}

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

   \begin{aligned} &=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{1-4 x^{2}}}}{\frac{-4 x}{\sqrt{1-4 x^{2}}}}=\frac{-1}{x}


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