Question

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

Answer 1

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

$={\tan }^{-1}\left(\frac{2\times 2^x}{1-(2^x{)}^2}\right)$

putting $2^x=\tan \theta$

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

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

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

$=2\theta$

$=2{\tan }^{-1}{2}^x$

$\frac{dy}{dx}=2\times \frac{1}{1+{\left(2^x\right)}^2}\times 2^x\log 2$

$=\frac{2\log {2.2}^x}{1+2^{2x}}$

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