cho hàm số y=cos3x.sin2x.y'( $\dfrac{\pi}{3}$ ) =? 13/09/2021 Bởi Alice cho hàm số y=cos3x.sin2x.y'( $\dfrac{\pi}{3}$ ) =?
Đáp án: $1$ Giải thích các bước giải: $y=\sin2x.\cos3x$ $=\dfrac{1}{2}\sin5x+\dfrac{1}{2}\sin(-x)$ $=\dfrac{1}{2}\sin5x-\dfrac{1}{2}\sin x$ $y’=\dfrac{5}{2}\cos 5x-\dfrac{1}{2}\cos x$ $\to y’\Big(\dfrac{\pi}{3}\Big)=\dfrac{5}{2}\cos\dfrac{5\pi}{3}-\dfrac{1}{2}\cos\dfrac{\pi}{3}=1$ Bình luận
Đáp án: $1$
Giải thích các bước giải:
$y=\sin2x.\cos3x$
$=\dfrac{1}{2}\sin5x+\dfrac{1}{2}\sin(-x)$
$=\dfrac{1}{2}\sin5x-\dfrac{1}{2}\sin x$
$y’=\dfrac{5}{2}\cos 5x-\dfrac{1}{2}\cos x$
$\to y’\Big(\dfrac{\pi}{3}\Big)=\dfrac{5}{2}\cos\dfrac{5\pi}{3}-\dfrac{1}{2}\cos\dfrac{\pi}{3}=1$
y’=(cos3x)’.sin2x+(sin2x)’.cos3x
=-3.sin3x.sin2x+2.cos2x.cos3x
=>y'(`\pi/3`)=1
ok a