Ответ:

[tex]\boxed{f'(x_{0}) = f' \bigg (\dfrac{\pi }{6} \bigg) =2(1 + 3\sqrt{3})}[/tex]

Пошаговое объяснение:

[tex]f(x) = 2x - \dfrac{3}{\sin x}[/tex]

[tex]x_{0} = \dfrac{\pi }{6}[/tex]

[tex]f'(x) = \bigg (2x - \dfrac{3}{\sin x} \bigg)' = \bigg (2x \bigg)' - \bigg ( \dfrac{3}{\sin x} \bigg)' = 2 - \dfrac{3' \cdot \sin x - 3 \cdot(\sin x)'}{\sin^{2}x} =[/tex]

[tex]= 2 - \dfrac{0 \cdot \sin x - 3 \cdot \cos x}{\sin^{2}x} = 2 + \dfrac{3 \cos x}{\sin^{2} x}[/tex]

[tex]f'(x_{0}) = f' \bigg (\dfrac{\pi }{6} \bigg) = 2 + \dfrac{3 \cos \bigg (\dfrac{\pi }{6} \bigg)}{\sin^{2} \bigg (\dfrac{\pi }{6} \bigg)} = 2 + \dfrac{3 \cdot \dfrac{\sqrt{3} }{2} }{\bigg (\dfrac{1}{2} \bigg)^{2} } = 2 + \dfrac{\dfrac{3\sqrt{3} }{2} }{\dfrac{1}{4}} = 2 + \dfrac{4 \cdot 3\sqrt{3}}{2 \cdot 1} =[/tex]

[tex]=2 + 2 \cdot 3\sqrt{3} = 2(1 + 3\sqrt{3})[/tex]