Find dy/dx : y= cos ^-1 (2^x+1/1+4^x)

Here, we need to find $\frac{d}{dx}\left({\cos }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)\right)$.
Now,
$$\begin{gathered} y={\cos }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \\ y={\cos }^{-1}\left(\frac{2.2^x}{1+{\left(2^x\right)}^2}\right) \end{gathered}$$

Substitute $2^x=\tan \theta$in the above equation.

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

Now, using the property $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$, we get
$$\begin{gathered} y={\cos }^{-1}\left(\sin 2\theta \right) \\ y={\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{2}-2\theta \right) \\ y=\frac{\mathrm{\pi }}{2}-2\theta \end{gathered}$$

Next, substituting $\theta ={\tan }^{-1}{2}^x$, we get
$y=\frac{\mathrm{\pi }}{2}-\left(2\right){\tan }^{-1}\left(2^x\right)$

Differentiate both sides with respect to x using the chain rule.
$$\begin{gathered} \frac{dy}{dx}=0-2\left(\frac{d}{dx}\left({\tan }^{-1}{2}^x\right)\right).\left(\frac{d}{dx}\left(2^x\right)\right) \\ \frac{dy}{dx}=-2\left(\frac{1}{1+{\left(2^x\right)}^2}\right).\left(2^x.\log 2\right) \\ \frac{dy}{dx}=-\left(\frac{2^{x+1}.\log 2}{1+4^x}\right) \end{gathered}$$

Therefore, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{-}\left(\frac{{\mathbf{2}}^{\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{.}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{2}}{\mathbf{1}\mathbf{+}{\mathbf{4}}^{\mathbf{x}}}\right)$