#### Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 22 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{-1}{1+x^{2}}$

Hint:

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

Given:

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

Solution:

Let,

$y=\sin ^{-1}\left \{ \frac{1}{\sqrt{1+x^{2}}} \right \}$

Let

$x=\cot \theta$

$\theta =\cot ^{-1}x$

Now,

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

Using,

\begin{aligned} &1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta \\ &y=\sin ^{-1}\left\{\frac{1}{\sqrt{\operatorname{cosec}^{2} \theta}}\right\} \\ &y=\sin ^{-1}\left\{\frac{1}{\operatorname{cosec} \theta}\right\} \end{aligned}                                                                                                    $\left \{ \frac{1}{cosec\theta }=\sin \theta \right \}$

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

$y=\theta$

$y=\cot ^{-1}x$

Differentiating with respect to x , We get

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