<s>[INST] Derivative of the function $\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$ 
Find y' if $y=\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$

In my reference $y'$ is given as $\frac{2^{x+1}\log2}{(1+4^x)}$. But is it a complete solution ?
Attempt 1
Let $2^x=\tan\alpha$
$$
\begin{align}
y=\sin^{-1}\Big(\frac{2\tan\alpha}{1+\tan^2\alpha}\Big)=\sin^{-1}(\sin2\alpha)&\implies \sin y=\sin2\alpha=\sin\big(2\tan^{-1}2^x\big)\\
&\implies y=n\pi+(-1)^n(2\tan^{-1}2^x)
\end{align}
$$
$$
\begin{align}
y'&=\pm\frac{2.2^x.\log2}{1+4^x}=\pm\frac{2^{x+1}.\log2}{1+4^x}\\
&=\color{blue}{\begin{cases}
\frac{2^{x+1}.\log2}{1+4^x}\text{ if }-n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq -n\pi+\frac{\pi}{2}\\
-\frac{2^{x+1}.\log2}{1+4^x}\text{ if }n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq n\pi+\frac{\pi}{2}
\end{cases}}
\end{align}
$$
Attempt 2
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-\frac{(2^{x+1})^2}{(1+4^x)^2}}}.\frac{d}{dx}\frac{2^{x+1}}{1+4^x}\\
&=\frac{1+4^x}{\sqrt{1+4^{2x}+2.4^x-4^x.4}}.\frac{(1+4^x)\frac{d}{dx}2^{x+1}-2^{x+1}\frac{d}{dx}(1+4^x)}{(1+4^x)^2}\\
&=\frac{(1+4^x).2^{x+1}.\log2-2^{x+1}.4^x.\log2.2}{\sqrt{1+4^{2x}-2.4^x}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1+4^x-2.4^x\big]}{\sqrt{(1-4^x)^2}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1-4^x\big]}{|{(1-4^x)}|.(1+4^x)}\\
&=\color{blue}{\begin{cases}\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1>4^x>0\\
-\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1<4^x
\end{cases}}
\end{align}
$$
In both my attempts i am getting both +ve and -ve solutions. Is it the right way to find the derivative?
And how do I connect the domain for each cases in attempt 1 and attempt 2 ?
 [/INST] $$y=\sin^{-1} \Big(\frac{2^{x+1}}{1+4^x}\Big)$$ the right side of the equation will be called $b$ it will help later. Take sine of both sides and the derivative $$\cos(y) y' =\frac{ln(2)2^{x+1}}{4^x+1}-\frac{ln(4) 4^x 2^{x+1}}{(4^x+1)^2}$$ the right side of the equation will be called $b'$ so $$ y' cos(y)=b'$$ divide both sides by cosine $$y'=\frac{b'}{\cos(y)}$$ substitute $y$ for $b$ and you have $$y'=\frac{b'}{cos(b)}$$
</s>