# Differentiate the following functions with respect to x:

Question:

Differentiate the following functions with respect to $x$ :

$\tan ^{-1}\left(\frac{2^{x+1}}{1-4^{x}}\right),-\infty Solution:$y=\tan ^{-1}\left\{\frac{2^{x+1}}{1-4^{x}}\right\}$Let$2^{x}=\tan \thetay=\tan ^{-1}\left\{\frac{2 \times 2^{x}}{1-\left(2^{x}\right)^{2}}\right\}y=\tan ^{-1}\left\{\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right\}$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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \tan ^{-1} 2^{x}\right)$

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

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