\(\begin{array}{l}g'\left( \varphi \right)\\ = \frac{{\left( {\cos \varphi + \sin \varphi } \right)'\left( {1 - \cos \varphi } \right) - \left( {\cos \varphi + \sin \varphi } \right)\left( {1 - \cos \varphi } \right)'}}{{{{\left( {1 - \cos \varphi } \right)}^2}}}\\ = \frac{{\left( { - \sin \varphi + \cos \varphi } \right)\left( {1 - \cos \varphi } \right) - \left( {\cos \varphi + \sin \varphi } \right)\left[ { - \left( { - \sin \varphi } \right)} \right]}}{{{{\left( {1 - \cos \varphi } \right)}^2}}}\\ = \frac{{ - \sin \varphi + \cos \varphi + \sin \varphi \cos \varphi - {{\cos }^2}\varphi - \sin \varphi \cos \varphi - {{\sin }^2}\varphi }}{{{{\left( {1 - \cos \varphi } \right)}^2}}}\\ = \frac{{ - \sin \varphi + \cos \varphi - \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)}}{{{{\left( {1 - \cos \varphi } \right)}^2}}}\\ = \frac{{ - \sin \varphi + \cos \varphi - 1}}{{{{\left( {1 - \cos \varphi } \right)}^2}}}\end{array}\) 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}
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