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

Hint:

$\frac{d}{dx}\left ( constant \right )=0$

$\frac{\partial }{\partial x}\left ( x^{n} \right )=nx^{n-1}$

Given:

$\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )$

Solution:

Let

$y=\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )$

To find the domain we need to find all $x$ such that

$-1 \leq \frac{2^{x+1}}{1+4 x} \leq 1$

Since the quantity in the middle is always positive, we need to find

all such that $\frac{2^{x+1}}{1+4 x} \leq 1$

ie, all $x$ such that $2^{x+1}=1+4^{x}$

$2\leq \frac{1}{2^{x}}+2^{x}$, which is true for a $x$

Hence the function is defined at all real numbers
Putting $2^{x}=\tan \theta$

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

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right) \\ &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right) \quad \text \; \; \; \; \; \; \; \; \; \; \; { Since, } \tan \theta=\frac{\sin \theta}{\cos \theta^{\prime} \sec \theta}, \frac{1}{\sec }=\cos \theta \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &\mathrm{U} \sin \mathrm{sin}=\sin 2 \theta=2 \sin \theta \cos \theta \end{aligned}

$y=\sin ^{-1}\left ( \sin \theta \right )$                                                                                                $\left\{\begin{array}{l} \sin ^{-1}(\sin \theta)=\theta \\ \text { if } \theta \in\left[\frac{-1}{2}, \frac{0}{2}\right] \end{array}\right\}$

$y=2\theta$

$y=2\tan ^{-1}\left ( 2^{x} \right )$                                                                                                      $\left\{\begin{array}{l} \text { since } 2^{\mathrm{x}} \quad=\tan \theta \\ \boldsymbol{\theta}=\tan ^{-1}\left(2^{\mathrm{x}}\right) \end{array}\right\}$

\begin{aligned} &\frac{d y}{d x} \Rightarrow \frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+\left(2^{x}\right)^{2}} \frac{\partial}{d x}\left(2^{x}\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+4^{x}} \cdot\left(2^{x}\right) \log 2 \\ &\frac{d y}{d x} \Rightarrow \frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}                                                        $\text { Since } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$

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