Cho \(y = {\tan ^3}x\). Tìm dy
Ta có:
\(y' = 3{\tan ^2}x.\frac{1}{{{{\cos }^2}x}} \Rightarrow dy = 3{\tan ^2}x.\frac{1}{{{{\cos }^2}x}}dx = \frac{{3{{\sin }^2}x}}{{{{\cos }^4}x}}\)
Chọn A.
-- Mod Toán 11