Differentiate the following functions with respect to $x$ :
$\tan ^{-1}\left(\frac{2^{x+1}}{1-4^{x}}\right),-\infty
$y=\tan ^{-1}\left\{\frac{2^{x+1}}{1-4^{x}}\right\}$
Let $2^{x}=\tan \theta$
$y=\tan ^{-1}\left\{\frac{2 \times 2^{x}}{1-\left(2^{x}\right)^{2}}\right\}$
$y=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\}$
Using $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$
$y=\tan ^{-1}(\tan 2 \theta)$
Considering the limits,
$-\infty $2^{-\infty}<2^{x}<2^{0}$ $0<\tan \theta<1$ $0<\theta<\frac{\pi}{4}$ $0<2 \theta<\frac{\pi}{2}$ Now, $y=\tan ^{-1}(\tan 2 \theta)$ $y=2 \theta$ $y=2 \tan ^{-1}\left(2^{x}\right)$ Differentiating w.r.t $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \tan ^{-1} 2^{x}\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}}=2 \times \frac{2^{\mathrm{x}} \log 2}{1+\left(2^{\mathrm{x}}\right)^{2}}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}+1} \log 2}{1+4^{\mathrm{x}}}$
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