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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 36 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 36 Maths Textbook Solution.

Answers (1)

Answer: Hence Prove ,\frac{dy}{dx}=\frac{2}{1+x^{2}}

Hint:

\begin{aligned} &\frac{d}{d x}(\text { constan } t)=0 ; \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

Given:

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{1+\mathrm{x}^{2}}}\right) \\ &0<\mathrm{x}<\infty \end{aligned}

Solution:

Let,

x=\tan \theta ,

\theta =\tan ^{-1}x

\begin{aligned} &y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^{2} \theta}}\right) \\ &y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{\sec ^{2} \theta}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{\sec ^{2} \theta}}\right) \\ &\text { Using } \sec ^{2} \theta=1+\tan ^{2} \theta \\ &\frac{\sin \theta}{\cos \theta}=\tan \theta \\ &\frac{1}{\sec \theta}=\theta \cos \theta \end{aligned}

\begin{aligned} &y=\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)+\cos ^{-1}\left(\frac{1}{\sec \theta}\right) \\ &y=\sin ^{-1}\left(\frac{\sin \theta}{\cos \theta} \times \cos \theta\right)+\cos ^{-1}(\cos \theta) \\ &y=\sin ^{-1}(\sin \theta)+\cos ^{-1}(\cos \theta)-(i) \end{aligned}

Considering limit

\begin{aligned} &0<\mathrm{x}<\infty \\ &0<\tan \theta<\infty \\ &0<\theta<\frac{\mathrm{\pi}}{2} \end{aligned}

So From eq (i)

y=\theta +\theta                                                                                                \text { Since, } \sin ^{-1}(\sin \theta)=\theta \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]

y=2\theta                                                                                                                    \cos ^{-1}(\cos \theta)=\theta \text { if } \theta \varepsilon[0, \pi]

y=2\tan ^{-1}x                                                                                            \left [ since x=\tan \theta \right ]

Differentiating it with respect to x

\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=2 \times \frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{2}{1+x^{2}} \end{aligned}

 

Posted by

infoexpert21

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