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#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 16 Maths Textbook Solution.

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

Hint:

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

Given:

\begin{aligned} &\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\} \\ &\frac{-1}{2}

Solution:

Let,

$y=\tan ^{-1}\left\{\frac{4 x}{1-4 x^{2}}\right\}$

Let,

$2x=\tan \theta$

$\theta =\tan ^{-1}2x$

$y =\tan ^{-1}\left \{ \frac{2\tan \theta }{1-\tan ^{2}\theta } \right \}$

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

$y=\tan ^{-1}\left ( \tan 2\theta \right )$

Considering the limits,

\begin{aligned} &\frac{-1}{2}                       $\left \{ \tan 45^{\circ}=1 \right \}$

Now,

$y=\tan ^{-1}\left ( \tan 2\theta \right )$

$y=2\theta$

$y=2\tan ^{-1}\left ( 2x \right )$                                                        $\left [ since,2x=\tan \theta \right ]$

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1} 2 x\right) \\ &\frac{d y}{d x}=2 \times \frac{2}{1+(2 x)^{2}} \\ &\frac{d y}{d x}=\frac{4}{1+4 x^{2}} \end{aligned}                            $\frac{d\left ( tan^{-1}x \right )}{dx}=\frac{1}{1+x^{2}}$