Question

найдите значение производной функции в заданной точке: f(x)=2x-3/sin x ,x=п/6​

Answer 1

Ответ:

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

Пошаговое объяснение:

$f(x) = 2x - \dfrac{3}{\sin x}$

$x_{0} = \dfrac{\pi }{6}$

$f'(x) = \bigg (2x - \dfrac{3}{\sin x} \bigg)' = \bigg (2x \bigg)' - \bigg ( \dfrac{3}{\sin x} \bigg)' = 2 - \dfrac{3' \cdot \sin x - 3 \cdot(\sin x)'}{\sin^{2}x} =$

$= 2 - \dfrac{0 \cdot \sin x - 3 \cdot \cos x}{\sin^{2}x} = 2 + \dfrac{3 \cos x}{\sin^{2} x}$

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

$=2 + 2 \cdot 3\sqrt{3} = 2(1 + 3\sqrt{3})$