Đáp án:
$y' =\dfrac{4\cos2x}{3\sqrt[3]{(1+2\sin2x)^2}}$
Giải thích các bước giải:
$y =\sqrt[3]{1 +2\sin2x}$
$\to y = (1 +2\sin2x)^{\tfrac13}$
$\to y' = \dfrac13\cdot(1 + 2\sin2x)^{\tfrac13 - 1}\cdot (1 +2\sin2x)'$
$\to y' = \dfrac13\cdot(1+2\sin2x)^{-\tfrac23}.2\cos2x.2$
$\to y' = \dfrac43\cos2x(1 +2\sin2x)^{-\tfrac23}$
$\to y' =\dfrac{4\cos2x}{3\sqrt[3]{(1+2\sin2x)^2}}$