If $f\left(x\right)=\frac{1+\tan x}{1-\tan x}$  then $f^{\mathrm{\prime }}\left(\frac{\pi }{6}\right)=$ 

If  $f\left(x\right)=\frac{1+\tan x}{1-\tan x}$ Then ,

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{\frac{d(1+\tan x)}{dx}\cdot (1-\tan x)-(1+\tan x)\cdot \frac{d(1-\tan x)}{dx}}{(1-\tan x{)}^2}\\ =\frac{\left(0+{\sec }^{2}x\right)(1-\tan x)-(1+\tan x)\left(0-{\sec }^{2}x\right)}{(1-\tan x{)}^2}\end{array}$

$\begin{array}{l}=\frac{2{\sec }^{2}x}{1+{\tan }^{2}x-2\tan x}=\frac{2{\sec }^{2}x}{{\sec }^{2}x-2\tan x}=\frac{2}{1-\sin 2x}\\ \therefore \text{ When }x=\frac{\pi }{6},f^{\mathrm{\prime }}\left(\frac{\pi }{6}\right)=\frac{2}{1-\sin \frac{\pi }{3}}=\frac{2}{1-\frac{\sqrt{3}}{2}}\\ =\frac{4}{2-\sqrt{3}}=4(2+\sqrt{3})\end{array}$