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Ответ:

Значение производной от функции [tex]\boldsymbol{g(x)}[/tex] в точке [tex]\boldsymbol{x_{0}}[/tex] равно [tex]\boldsymbol{-2,875}[/tex]

Примечание:

[tex]\bigg(\dfrac{f}{g} \bigg)' = \dfrac{f'g - fg'}{g^{2}}[/tex], где [tex]f,g \ -[/tex] функции от аргумента x.

Пошаговое объяснение:

[tex]x_{0} = 1[/tex]

[tex]g(x) = \dfrac{x^{3} - 4x^{2} }{\sqrt{x} +\dfrac{1}{x} }[/tex]

[tex]g'(x) = \Bigg( \dfrac{x^{3} - 4x^{2} }{\sqrt{x} +\dfrac{1}{x} } \Bigg)' = \dfrac{(x^{3} - 4x^{2})'\bigg(\sqrt{x} +\dfrac{1}{x} \bigg) - (x^{3} - 4x^{2})\bigg(\sqrt{x} +\dfrac{1}{x} \bigg)'}{\bigg(\sqrt{x} +\dfrac{1}{x} \bigg)^{2} } =[/tex]

[tex]=\dfrac{(3x^{2} - 8x)\bigg(\sqrt{x} +\dfrac{1}{x} \bigg) - (x^{3} - 4x^{2})\bigg(\dfrac{1}{2\sqrt{x} } -\dfrac{1}{x^{2} } \bigg)}{\bigg(\sqrt{x} +\dfrac{1}{x} \bigg)^{2} }[/tex]

[tex]g'(1) = \dfrac{(3 \cdot 1^{2} - 8\cdot 1)\bigg(\sqrt{1} +\dfrac{1}{1} \bigg) - (1^{3} - 4 \cdot 1^{2})\bigg(\dfrac{1}{2\sqrt{1} } -\dfrac{1}{1^{2} } \bigg)}{\bigg(\sqrt{1} +\dfrac{1}{1} \bigg)^{2} } =[/tex]

[tex]= \dfrac{(3 - 8)\bigg(1 +1 \bigg) - (1 - 4 )\bigg(\dfrac{1}{2} -\dfrac{1}{1 } \bigg)}{\bigg(1 +1 \bigg)^{2} } = \dfrac{-5 \cdot 2 - (-3) \cdot (-0,5)}{2^{2} } =[/tex]

[tex]= \dfrac{-10 - 1,5}{4 } = - \dfrac{11,5}{4} = -2,875[/tex]