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#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 19

Answer: $\frac{-2}{ax}$

Hint: $\text { Let } a x=\sin \theta$

Given: $\sin ^{-1}\left(2 a x \sqrt{1-a^{2} x^{2}}\right) \text { w.r.t } \sqrt{1-a^{2} x^{2}}$

$\frac{-1}{\sqrt{2}}

Explanation:

$\text { Let } \mathrm{a} x=\sin \theta$

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

Now

\begin{aligned} &\frac{-1}{\sqrt{2}}

\begin{aligned} &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\\\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}

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

\begin{aligned} &\frac{d u}{d x}=\frac{2}{\sqrt{1-a^{2} x^{2}}}(a) \\\\ &=\frac{2 a}{\sqrt{1-a^{2} x^{2}}} \end{aligned}

\begin{aligned} &v=\sqrt{1-a^{2} x^{2}} \\\\ &\frac{d v}{d x}=\frac{1}{2 \sqrt{1-a^{2} x^{2}}}\left(-a^{2} 2 x\right) \\\\ &\frac{d v}{d x}=\frac{-a^{2} x}{\sqrt{1-a^{2} x^{2}}} \end{aligned}$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2 a}{\sqrt{1-a^{2} x^{2}}}}{\frac{-a^{2} x}{\sqrt{1-a^{2} x^{2}}}}=\frac{-2}{a x}$