Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {\tan x} \right)' = \left( {\frac{{\sin x}}{{\cos x}}} \right)' = \frac{{\left( {\sin x} \right)'.\cos x - \left( {\cos x} \right)'.\sin x}}{{{{\cos }^2}x}}\\
= \frac{{\cos x.\cos x - \left( { - \sin x} \right).\sin x}}{{{{\cos }^2}x}} = \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} = \frac{1}{{{{\cos }^2}x}}\\
\Rightarrow \left( {\tan u} \right)' = \frac{{u'}}{{{{\cos }^2}u}}\\
a,\\
f\left( x \right) = 2\tan x \Rightarrow f'\left( x \right) = 2.\left( {\tan x} \right)' = \frac{2}{{{{\cos }^2}x}}\\
b,\\
f\left( x \right) = \tan \left( {x - \frac{{2\pi }}{3}} \right)\\
\Rightarrow f'\left( x \right) = \dfrac{{\left( {x - \frac{{2\pi }}{3}} \right)'}}{{{{\cos }^2}\left( {x - \frac{{2\pi }}{3}} \right)}} = \dfrac{1}{{{{\cos }^2}\left( {x - \frac{{2\pi }}{3}} \right)}}
\end{array}\)
