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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 6

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 6

Answers (1)

Answer: \frac{1}{4}

Hint:    \text { Let } u=\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right],

            v=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right]
Given: \tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right] \text { w.r.t } \sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right]

            -1<x<1, x \neq 0

Explanation:  \text { Let } x=\tan \theta

u=\tan ^{-1}\left[\frac{\sqrt{1+x^{2}}-1}{x}\right]

   \begin{aligned} &=\tan ^{-1}\left[\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right] \\\\ &=\tan ^{-1}\left[\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right] \\\\ &=\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right] \end{aligned}

   \begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right] \\\\ &=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] \end{aligned}     .......................(1)

\begin{aligned} &\text { Now } \cos 2 \theta=1-2 \sin ^{2} \theta \\\\ &2 \sin ^{2} \theta=1-\cos 2 \theta \\\\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \end{aligned}

   \begin{aligned} &\sin 2 \theta=2 \sin \theta \cos \theta \\\\ &\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{aligned}

Put in (1)

\begin{aligned} &u=\tan ^{-1}\left[\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \\\\ &u=\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \end{aligned}

Now,

\begin{aligned} &-1<x<1 \\\\ &-1<\tan \theta<1 \\\\ &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \end{aligned}    .........(2)

\begin{aligned} &-\frac{\pi}{8}<\frac{\theta}{2}<\frac{\pi}{8} \\\\ &u=\frac{\theta}{2} \text { as } \frac{\theta}{2} \in\left(-\frac{\pi}{8}, \frac{\pi}{8}\right) \end{aligned}

\begin{aligned} &u=\frac{\tan ^{-1} x}{2} \\\\ &\frac{d u}{d x}=\frac{1}{2}\left[\frac{1}{1+x^{2}}\right] \end{aligned}

       =\frac{1}{2\left(1+x^{2}\right)}

\begin{aligned} &v=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right] \\\\ &x=\tan \theta \\\\ &v=\sin ^{-1}\left[\frac{2(\tan \theta)}{1+\tan ^{2} \theta}\right] \end{aligned}

    =\sin ^{-1}(\sin 2 \theta)

-\frac{\pi}{4}<\theta<\frac{\pi}{4}    ........From (2)

-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}

v=\sin ^{-1}(\sin 2 \theta) \; \; \; \; \; \quad 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

    =2 \theta

\begin{aligned} &v=2 \tan ^{-1} x \\\\ &\frac{d v}{d x}=\frac{2}{1+x^{2}} \end{aligned}

\begin{aligned} &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{1+x^{2}}} \\\\ &=\frac{1}{4} \end{aligned}

 

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