$f\left(x\right)=[\mathrm{Tan}\frac{\mathrm{\pi }}{4}+\mathrm{Tan}x]\cdot [\mathrm{Tan}\frac{\mathrm{\pi }}{4}+\mathrm{Tan}(\frac{\mathrm{\pi }}{4}-x\left)\right]{\text{ and g(x)=x}}^2\text{+1 Then the value }$

$\text{ of }{g}^1\left[f\right(x\left)\right]\text{ is equals to }$

$\begin{array}{l}f\left(x\right)=(1+\tan x)\left(1+\frac{1-\tan x}{1+\tan x}\right)\\ =(1+\tan x)\left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)\\ f\left(x\right)=2\\ g\left(x\right)=x^2+x\\ g^1\left(x\right)=2x+1\\ g^1\left(f\right(x\left)\right)=g^1\left(2\right)\\ =2\left(2\right)+1\\ =5\end{array}$