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