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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 10

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 10

Answers (1)

Answer: \frac{1}{a x \sqrt{1+a^{2} x^{2}}}

Hint:   \text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right), v=\sqrt{1+a^{2} x^{2}}
Given:  \tan ^{-1}\left(\frac{1+a x}{1-a x}\right) \text { w.r.t } \sqrt{1+a^{2} x^{2}}

Explanation:  \text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)

\begin{aligned} &v=\sqrt{1+a^{2} x^{2}} \\\\ &\text { Let } a x=\tan \theta \\\\ &u=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \end{aligned}

   =\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right) \quad\left[\tan \frac{\pi}{4}=1\right]

   \begin{aligned} &=\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right) \\\\ &=\frac{\pi}{4}+\theta \end{aligned}

u=\frac{\pi}{4}+\tan ^{-1}(a x) \; \; \; \; \; \; \; \quad\left[\begin{array}{l} a x=\tan \theta \\ \theta=\tan ^{-1} a x \end{array}\right]

\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+(a x)^{2}}[a \times 1] \\\\ &=\frac{a}{1+a^{2} x^{2}} \\\\ &v=\sqrt{1+a^{2} x^{2}} \end{aligned}

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

 

 

Posted by

infoexpert26

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