#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 18 Maths Textbook Solution.

Answer:$\frac{d y}{d x}=\frac{2 a^{x} l o g a}{1+a^{2 x}}$

Hint:

$\frac{d c}{d x}=0;\frac{d}{dx}(x^{n})=nx^{n-1}$

Given:

$\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\}, \mathrm{a}>1,-\infty<\mathrm{x}<0$

Solution:

Let

$y=\tan ^{-1}\left\{\frac{2 \mathrm{a}^{x}}{1-\mathrm{a}^{2 x}}\right\},$

Let,

\begin{aligned} &\begin{aligned} &\mathrm{a}^{\mathrm{x}}=\tan \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\} \\ &\text { Using } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \\ &\mathrm{y}=\tan ^{-1}(\tan 2 \theta) \end{aligned}\\ &\text { Considering the limits, }\\ &-\infty<\mathrm{x}<0\\ &a^{-\infty}

$0< \tan \theta < 1$                                                                                                                $\left \{ a^{o} =1\right \}$

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

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

$y=2\theta$

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

Differentiating with respect to $x$, we get

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