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

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

Answers (1)

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}}<a x<\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}}<a x<\frac{1}{\sqrt{2}} \\\\ &\frac{-1}{\sqrt{2}}<\sin \theta<\frac{1}{\sqrt{2}} \end{aligned}

\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}

 

Posted by

infoexpert26

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