Please Solve R.D. Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 12 Sub Question 1 Maths Textbook Solution.

$\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}$

Hint:

$\frac{d}{dx}(constant)=0$

$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

Given:

$\tan ^{-1}\left \{ \frac{x}{1+\sqrt{1-x^{2}}} \right \}$

$-1< x< 1$

Solution:

Let,

\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\frac{\mathrm{x}}{1+\sqrt{1-\mathrm{x}^{2}}}\right\} \\ &\mathrm{x}=\sin \theta \\ &\theta=\sin ^{-1} \mathrm{x} \end{aligned}

Now,$y=\tan ^{-1}\left \{ \frac{\sin \theta }{1+\sqrt{1\sin ^{2}\theta }} \right \}$

Using $\sin ^{2}\theta +\cos ^{2}\theta =1$

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

\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right\} \\ &\tan \theta=\frac{\sin \theta}{\cos \theta} \\ &\mathrm{y}=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \end{aligned}

Considering the limits,

\begin{aligned} &-1

Now,

\begin{aligned} &\mathrm{y}=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \\ &\mathrm{y}=\frac{\theta}{2} \end{aligned}                                $\tan ^{-1}(\tan \theta)=0 \text { if } \theta \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

$y=\frac{1}{2}\sin ^{-1}x$

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=-1 \\ &\frac{d}{d x}(\text { constants })=0 \\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} x\right)\right. \\ &\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \end{aligned}

$-1< x< 1$