Cho hàm số $f\left( x \right) = \tan \left( {x - \dfrac{{2\pi }}{3}} \right)$. Giá trị $f'\left( 0 \right)$ bằng:

$\begin{array}{l}f\left( x \right) = \tan \left( {x - \dfrac{{2\pi }}{3}} \right) = \dfrac{{\tan x - \tan \dfrac{{2\pi }}{3}}}{{1 + \tan x.\tan \dfrac{{2\pi }}{3}}} = \dfrac{{\tan x + \sqrt 3 }}{{1 - \sqrt 3 \tan x}}\\f'\left( x \right) = \dfrac{{\left( {\tan x + \sqrt 3 } \right)'\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan x} \right)'}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{{\dfrac{1}{{{{\cos }^2}x}}\left( {1 - \sqrt 3 \tan x} \right) - \left( {\tan x + \sqrt 3 } \right)\left( { - \dfrac{{\sqrt 3 }}{{{{\cos }^2}x}}} \right)}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{{\dfrac{1}{{{{\cos }^2}x}} - \dfrac{{\sqrt 3 \tan x}}{{{{\cos }^2}x}} + \dfrac{{\sqrt 3 \tan x}}{{{{\cos }^2}x}} + \dfrac{3}{{{{\cos }^2}x}}}}{{{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\f'\left( x \right) = \dfrac{4}{{{{\cos }^2}x{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\\ \Rightarrow f'\left( 0 \right) = \dfrac{4}{{1\left( {1 - \sqrt 3 .0} \right)}} = 4\end{array}$