Đáp án:
$a) y'=2cos\left ( 2x+\frac{\pi}{6} \right )+1\\
b) y'=cos(2-3x)+3xsin(2-3x)$
Giải thích các bước giải:
$a) y=sin\left ( 2x+\frac{\pi}{6} \right )+x\\
\Rightarrow y'=\left [ sin\left ( 2x+\frac{\pi}{6} \right )+x\right ]'\\
=(2x+\frac{\pi}{6})'cos\left ( 2x+\frac{\pi}{6} \right )+1\\
=2cos\left ( 2x+\frac{\pi}{6} \right )+1\\
b) y=x.cos(2-3x)\\
\Rightarrow y'=\left [ x.cos(2-3x) \right ]'\\
=x'.cos(2-3x)+x.cos(2-3x)'\\
=cos(2-3x)-x.(2-3x)'.sin(2-3x)\\
=cos(2-3x)-x.(-3).sin(2-3x)\\
=cos(2-3x)+3xsin(2-3x)$
