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  • Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 39 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 10  Differentiation  Exercise 10.3 Question 39 Maths Textbook Solution.

Answers (1)

Answer: \frac{dy}{dx}=\frac{4}{1+x^{2}}

Hint:

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

Given:

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

Prove

\frac{dy}{dx}=\frac{1}{1+x^{2}}

Solution:

Here,y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) \

Using \sec ^{-1}

      \begin{aligned} &\mathrm{x}=\cos ^{-1}\left(\frac{1}{\mathrm{x}}\right) \\ &\mathrm{y}=\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)+\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \end{aligned}

Put x=\tan \theta

\begin{aligned} &y=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\ &y=\tan ^{-1}(\tan 2 \theta)+\cos ^{-1}(\cos 2 \theta)-(1) \end{aligned}

Using,

        \tan 2\theta =\frac{2\tan \theta }{1-\tan ^{2}\theta }

\begin{aligned} &\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta \\ &\text { Here, } 0<\mathrm{x}<\infty \\ &0<\tan \theta<\infty \\ &0<\theta<\frac{\mathrm{\pi}}{2} \\ &0<2 \theta<\mathrm{\pi} \end{aligned}

So From eq(i)

y=2\theta +2\theta

\left\{\text { Since, } \tan ^{-1}(\tan \theta)=\theta \text { if } \theta \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text { and } \cos ^{-1}(\cos \theta)=\theta \text { if } \theta[0, \pi]\right\}

y=4\theta

y=4\tan ^{-1}x                                                                                                            \left [ Using\: x=\tan \theta ,\theta =\tan ^{-1}x \right ]

Differentiating it with respect t0 x

\begin{aligned} &\frac{d y}{d x}=\frac{4}{1+ x^{2}} \\ &\Rightarrow \frac{4}{1+x^{2}} \\ &\text { Using } \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}} \end{aligned}

 

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