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Câu hỏi: Tìm đạo hàm của hàm số sau:
\(g\left( \varphi \right) = {{\cos \varphi + \sin \varphi } \over {1 - \cos \varphi }}.\)
\(g\left( \varphi \right) = {{\cos \varphi + \sin \varphi } \over {1 - \cos \varphi }}.\)
Lời giải chi tiết
\(\begin{array}{l}
g'\left(\varphi \right)\\
= \frac{{\left({\cos \varphi + \sin \varphi } \right)'\left({1 - \cos \varphi } \right) - \left({\cos \varphi + \sin \varphi } \right)\left({1 - \cos \varphi } \right)'}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{\left({ - \sin \varphi + \cos \varphi } \right)\left({1 - \cos \varphi } \right) - \left({\cos \varphi + \sin \varphi } \right)\left[ { - \left({ - \sin \varphi } \right)} \right]}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi + \sin \varphi \cos \varphi - {{\cos }^2}\varphi - \sin \varphi \cos \varphi - {{\sin }^2}\varphi }}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi - \left({{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi - 1}}{{{{\left({1 - \cos \varphi } \right)}^2}}}
\end{array}\)
\(\begin{array}{l}
g'\left(\varphi \right)\\
= \frac{{\left({\cos \varphi + \sin \varphi } \right)'\left({1 - \cos \varphi } \right) - \left({\cos \varphi + \sin \varphi } \right)\left({1 - \cos \varphi } \right)'}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{\left({ - \sin \varphi + \cos \varphi } \right)\left({1 - \cos \varphi } \right) - \left({\cos \varphi + \sin \varphi } \right)\left[ { - \left({ - \sin \varphi } \right)} \right]}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi + \sin \varphi \cos \varphi - {{\cos }^2}\varphi - \sin \varphi \cos \varphi - {{\sin }^2}\varphi }}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi - \left({{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)}}{{{{\left({1 - \cos \varphi } \right)}^2}}}\\
= \frac{{ - \sin \varphi + \cos \varphi - 1}}{{{{\left({1 - \cos \varphi } \right)}^2}}}
\end{array}\)