Đáp án:
\[y' = \cos \left( {x + \pi } \right) - \left( {2x - 2} \right).\sin \left( {{x^2} - 2x + \frac{\pi }{3}} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \sin \left( {x + \pi } \right) + \cos \left( {{x^2} - 2x + \frac{\pi }{3}} \right)\\
\Rightarrow y' = \left( {x + \pi } \right)'.\cos \left( {x + \pi } \right) + \left( {{x^2} - 2x + \frac{\pi }{3}} \right)'.\left( { - \sin \left( {{x^2} - 2x + \frac{\pi }{3}} \right)} \right)\\
= 1.\cos \left( {x + \pi } \right) + \left( {2x - 2} \right).\left( { - \sin \left( {{x^2} - 2x + \frac{\pi }{3}} \right)} \right)\\
= \cos \left( {x + \pi } \right) - \left( {2x - 2} \right).\sin \left( {{x^2} - 2x + \frac{\pi }{3}} \right)
\end{array}\)
Vậy \(y' = \cos \left( {x + \pi } \right) - \left( {2x - 2} \right).\sin \left( {{x^2} - 2x + \frac{\pi }{3}} \right)\)
