Cho \({f\left( x \right) = \tan \left( {2{x^3} - 5} \right)}\). Tìm
Ta có:
\(f'\left( x \right) = \frac{{{{\left( {2{x^3} - 5} \right)}^\prime }}}{{{{\cos }^2}\left( {2{x^3} - 5} \right)}} = \frac{{6{x^2}}}{{{{\cos }^2}\left( {2{x^3} - 5} \right)}}\)
Chọn D.
-- Mod Toán 11