Question

giúp mình với cả nhà ơi

Answer 1

a.

\(y'=cos\left(3x+\dfrac{\pi}{2}\right).\left(3x+\dfrac{\pi}{2}\right)'=3cos\left(3x+\dfrac{\pi}{2}\right)=-3sin3x\)

b.

\(y'=-sin\left(3x-\dfrac{\pi}{6}\right).\left(3x-\dfrac{\pi}{6}\right)'=-3sin\left(3x-\dfrac{\pi}{6}\right)\)

Answer 2

a,\(y=sin\left(3x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow\left(3x+\dfrac{\pi}{2}\right)'cos\left(3x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow3cos\left(3x+\dfrac{\pi}{2}\right)\)

b,\(y=cos\left(3x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow-\left(3x-\dfrac{\pi}{6}\right)'sin\left(3x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow-3sin\left(3x-\dfrac{\pi}{6}\right)\)