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Câu hỏi: Tìm đạo hàm của hàm số sau:
\(y = {2 \over {\cos \left( {{\pi \over 6} - 5x} \right)}}.\)
\(y = {2 \over {\cos \left( {{\pi \over 6} - 5x} \right)}}.\)
Lời giải chi tiết
\(\begin{array}{l}
y' = \dfrac{{ - 2\left[ {\cos \left({\dfrac{\pi }{6} - 5x} \right)} \right]'}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{ - 2.\left({\dfrac{\pi }{6} - 5x} \right)'\left[ { - \sin \left({\dfrac{\pi }{6} - 5x} \right)} \right]}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{2.\left({ - 5} \right)\sin \left({\dfrac{\pi }{6} - 5x} \right)}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{ - 10\sin \left({\dfrac{\pi }{6} - 5x} \right)}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}
\end{array}\)
\(\begin{array}{l}
y' = \dfrac{{ - 2\left[ {\cos \left({\dfrac{\pi }{6} - 5x} \right)} \right]'}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{ - 2.\left({\dfrac{\pi }{6} - 5x} \right)'\left[ { - \sin \left({\dfrac{\pi }{6} - 5x} \right)} \right]}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{2.\left({ - 5} \right)\sin \left({\dfrac{\pi }{6} - 5x} \right)}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}\\
= \dfrac{{ - 10\sin \left({\dfrac{\pi }{6} - 5x} \right)}}{{{{\cos }^2}\left({\dfrac{\pi }{6} - 5x} \right)}}
\end{array}\)