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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 37 Sub Question 2 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 37 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{1}{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:

\cos ^{-1}\left ( \frac{1-x}{1+x} \right )

Solution:

y=\cos ^{-1}\left ( \frac{1-x}{1+x} \right )

Let,

\begin{aligned} &\mathrm{x}=\tan \theta \\ &\theta=\tan ^{-1} \mathrm{x} \\ &\mathrm{y}=\cot ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ &\mathrm{y}=\cot ^{-1}\left(\frac{\tan \frac{\mathrm{n}}{4}-\tan \theta}{1+\tan \frac{\mathrm{n}}{4} \tan \theta}\right) \end{aligned}

\begin{aligned} &y=\cot ^{-1}\left[\tan \left(\frac{n}{4}-\theta\right)\right]\\ &\text { Using, }\\ &\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\\ &y=\cot ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]\\ &\mathrm{y}=\cot ^{-1}\left[\operatorname{cots}\left(\frac{\mathrm{\pi}}{2}-\left(\frac{\mathrm{\pi}-\theta}{4}-\theta\right)\right]\right. \end{aligned}                                        \therefore since\: \cot \left ( \frac{\pi}{2}-\theta \right )=\tan \theta

\begin{aligned} &y=\cot ^{-1}\left[\cot \left(\frac{\pi}{2}-\frac{\pi}{4}+\theta\right)\right] \\ &y=\cot ^{-1}\left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\ &y=\frac{\pi}{4}+\theta \end{aligned}                                        \left\{\cot ^{-1}(\cot \theta)=\theta, \text { if } \theta<\left[-\frac{\mathbf{\pi}}{2}, \frac{\pi}{2}\right]\right\}

y=\frac{\pi}{4}+\tan ^{-1}x                                                                                                    \left\{\begin{array}{c} \text { Since } \mathrm{x}=\tan \theta \\ \theta=\tan ^{-1} \mathrm{x} \end{array}\right\}

Differentiating it with respect to x,

\begin{aligned} &\frac{d y}{d x}=0+\frac{1}{1+x^{2}} \\ &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{\partial}{d x}(\text { constant })=0 \\ &\frac{d y}{d x}=\frac{1}{1+x^{2}} \end{aligned}

 

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