If $\mathrm{y}={\tan }^{-1} \left(\frac{2^{\mathrm{x}}}{1+2^{2\mathrm{x}+1}}\right)$ then $\frac{dy}{dx}$at x= 0 is

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

$$\begin{gathered} \begin{array}{rr}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+{\left(2^{\mathrm{x}+1}\right)}^2}\cdot 2^{\mathrm{x}+1}\log 2-\frac{1}{1+{\left(2^{\mathrm{x}}\right)}^2}{2}^{\mathrm{x}}\log 2 \\ =\frac{2\log 2}{1+4}-\frac{\log 2}{1+1} \\ =\log 2\left(\frac{2}{5}-\frac{1}{2}\right)=\frac{-1}{10}\log 2& \end{array} \end{gathered}$$$\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{x}=0$