Question

If $y={\sin }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$, find $\frac{dy}{dx}$

Answer 1

$\begin{array}{c}y={\sin }^{-1}\left(\frac{2^x+1}{1+4^x}\right)\\ y={\sin }^{-1}\left(\frac{{2.2}^x}{1+{\left(2^x\right)}^2}\right)\\ ={\sin }^{-1}\left(\frac{2t}{1+t^2}\right).............\left[t=2^x\right]\\ =2{\tan }^{-1}tx\in (-\infty ,0]\\ =\pi -2{\tan }^{-1}x>0\\ y=2{\tan }^{-1}\left(2^x\right)x\in (-\infty ,0]\\ =\pi -2{\tan }^{-1}x>0\\ \frac{dy}{dx}=\frac{2\left(2^x\right){\ln }_2}{1+4^x}\\ forx\in (-\infty ,0]\end{array}$

$\frac{dy}{dx}=\frac{-22^x{\ln }_2}{1+4^x}x>0$