Đáp án: $1$
Giải thích các bước giải:
$y=\sin2x.\cos3x$
$=\dfrac{1}{2}\sin5x+\dfrac{1}{2}\sin(-x)$
$=\dfrac{1}{2}\sin5x-\dfrac{1}{2}\sin x$
$y'=\dfrac{5}{2}\cos 5x-\dfrac{1}{2}\cos x$
$\to y'\Big(\dfrac{\pi}{3}\Big)=\dfrac{5}{2}\cos\dfrac{5\pi}{3}-\dfrac{1}{2}\cos\dfrac{\pi}{3}=1$