\(\begin{array}{l}a)\,\,y = 5\sin x - 3\cos x\\ \Rightarrow y' = 5\cos x + 3\sin x\\b)\,\,y = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\\ \Rightarrow y' = \dfrac{{\left( {\cos x - \sin x} \right)\left( {\sin x - \cos x} \right) - \left( {\sin x + \cos x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\\,\,\,\,\,y' = \dfrac{{2\sin x\cos x - 1 - 1 - 2\sin x\cos x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\\,\,\,\,\,y' = \dfrac{{ - 2}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\c)\,\,y = x\cot x\\ \Rightarrow y' = \cot x - \dfrac{x}{{{{\sin }^2}x}}\\d)\,\,y = \dfrac{{\sin x}}{x} + \dfrac{x}{{\sin x}}\\ \Rightarrow y' = \dfrac{{x\cos x - \sin x}}{{{x^2}}} + \dfrac{{\sin x - x\cos x}}{{{{\sin }^2}x}}\\\,\,\,\,\,\,y' = \left( {x\cos x - \sin x} \right)\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{{\sin }^2}x}}} \right)\\e)\,\,y = \sqrt {1 + 2\tan x} \\ \Rightarrow y' = \dfrac{{\dfrac{2}{{{{\cos }^2}x}}}}{{2\sqrt {1 + 2\tan x} }}\\\,\,\,\,\,\,\,y' = \dfrac{1}{{{{\cos }^2}x.\sqrt {1 + 2\tan x} }}\\f)\,\,y = \sin \sqrt {1 + {x^2}} \\ \Rightarrow y' = \cos \sqrt {1 + {x^2}} .\dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}\\\,\,\,\,\,\,y' = \dfrac{x}{{\sqrt {1 + {x^2}} }}\cos \sqrt {1 + {x^2}} \end{array}\)
