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

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}