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

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}