#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Diffrentiation Exercise 10.3 Question 17 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{2^{x+1}log2}{1+4x}$

Hint:

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

Given:

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

Solution:

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

Let,

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

$y=\tan ^{-1}\left \{ \frac{2x2^{x}}{1-\left ( 2x \right )^{2}} \right \}$

As we Know

\begin{aligned} &a^{m+n} \rightarrow a^{m} \cdot a^{n} \\ &y=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \end{aligned}

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} &-\infty                                                    $\left\{2^{0}=1,2^{-\infty}=0\right\}$

$0< 2\theta < \frac{\pi}{2}$                                                                $\left \{ \tan \frac{\pi}{4} =1\right \}$

$0< 2\theta < \frac{\pi}{2}$

Now

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

$y==2\theta$

$y=2\tan ^{-1}\left ( 2^{x} \right )$                                                                                                         $\left\{\begin{array}{r} \sin c e 2^{x}=\tan \theta \\ \theta=\tan ^{-1}\left(2^{x}\right) \end{array}\right\}$

Differentiating with respect to x , We get

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{2^{x} \log 2}{1+\left(2^{x}\right)^{2}} \\ &\frac{d y}{d x}=\frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}                                                                            $\begin{gathered} \therefore \frac{\mathrm{d}\left(\tan ^{-1} \mathbf{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{\mathrm{x}}\right)=\log 2 \\ \left\{\mathrm{a}^{\mathrm{m}} \cdot \mathbf{a}^{\mathrm{n}} \Rightarrow \mathbf{a}^{\mathrm{m}+\mathrm{h}}\right\} \end{gathered}$