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

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

Answers (1)

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}

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