Differentiate the function $\sin^{-1}\left( \frac{2^{x+1}}{1 + 4^x} \right)$ with respect to $x$.

$\frac{2^{x+1} \ln 2}{1 + 4^x}$

The correct answer is Option (3) → $\frac{2^{x+1} \ln 2}{1 + 4^x}$ ##

Let $f(x) = \sin^{-1} \left( \frac{2^{x+1}}{1+4^x} \right)$. To find the domain of this function we need to find all $x$ such that $-1 \le \frac{2^{x+1}}{1+4^x} \le 1$. Since the quantity in the middle is always positive, we need to find all $x$ such that $\frac{2^{x+1}}{1+4^x} \le 1$, i.e., all $x$ such that $2^{x+1} \le 1+4^x$. We may rewrite this as $2 \le \frac{1}{2^x} + 2^x$ which is true for all $x$. Hence the function is defined at every real number. By putting $2^x = \tan \theta$, this function may be rewritten as

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

$= \sin^{-1} \left[ \frac{2^x \cdot 2}{1 + (2^x)^2} \right] $

$= \sin^{-1} \left[ \frac{2 \tan \theta}{1 + \tan^2 \theta} \right] $

$= \sin^{-1} [\sin 2\theta]$

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

Thus $f'(x) = 2 \cdot \frac{1}{1+(2^x)^2} \cdot \frac{d}{dx}(2^x)$

$= \frac{2}{1+4^x} \cdot (2^x) \log 2 $

$= \frac{2^{x+1} \log 2}{1+4^x}$