Differentiate the following functions with respect to x :

Solution:

$y=\sin ^{-1}\left\{\frac{2^{x+1}}{1+4^{x}}\right\}$

For function to be defined

$-1 \leq \frac{2^{x+1}}{1+4^{x}} \leq 1$

Since the quantity is positive always

$\Rightarrow 0 \leq \frac{2^{x+1}}{1+4^{x}} \leq 1$

$\Rightarrow 0<2^{x+1} \leq 1+4^{x}$

$\Rightarrow 0<2 \leq 2^{-x}+2^{x}$

This condition is always true, hence function is always defined.

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

Let $2^{x}=\tan \theta$

$y=\sin ^{-1}\left\{\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right\}$

Using $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

Now,

$y=\sin ^{-1}(\sin 2 \theta)$

$y=2 \theta$

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

Differentiating w.r.t $\mathrm{x}$, we get

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

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

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