Ответ:

Производная функции:

[tex]\boldsymbol{\boxed{f'(x) = \dfrac{1}{\sqrt{x} } }}[/tex]

Пошаговое объяснение:

Производная по определению:

[tex]\displaystyle y'(x_{0}) = \lim_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \dfrac{y(x_{0} + \Delta x) - y(x_{0})}{\Delta x}[/tex]

[tex]f(x) = \sqrt{4x}[/tex]

[tex]f(x_{0} + \Delta x) = \sqrt{4(x_{0} + \Delta x)}[/tex]

[tex]\displaystyle f'(x_{0}) = \lim_{\Delta x \to 0} \dfrac{ \sqrt{4(x_{0} + \Delta x)} - \sqrt{4x_{0}} }{\Delta x} = \lim_{\Delta x \to 0} \dfrac{ 2\sqrt{(x_{0} + \Delta x)} - 2\sqrt{x_{0}} }{\Delta x} =[/tex]

[tex]\displaystyle =2 \lim_{\Delta x \to 0} \dfrac{ \sqrt{(x_{0} + \Delta x)} - \sqrt{x_{0}} }{\Delta x} = 2 \lim_{\Delta x \to 0} \dfrac{(\sqrt{(x_{0} + \Delta x)} - \sqrt{x_{0}}) (\sqrt{(x_{0} + \Delta x)} + \sqrt{x_{0}}) }{\Delta x (\sqrt{(x_{0} + \Delta x)} + \sqrt{x_{0}})} =[/tex]

[tex]\displaystyle = 2 \lim_{\Delta x \to 0} \dfrac{x_{0} + \Delta x - x_{0}}{\Delta x (\sqrt{(x_{0} + \Delta x)} + \sqrt{x_{0}})} = 2 \lim_{\Delta x \to 0} \dfrac{ \Delta x }{\Delta x (\sqrt{(x_{0} + \Delta x)} + \sqrt{x_{0}})} =[/tex]

[tex]\displaystyle= 2 \lim_{\Delta x \to 0} \dfrac{1 }{\sqrt{(x_{0} + \Delta x)} + \sqrt{x_{0}}} = \frac{2}{2\sqrt{x_{0}} } = \frac{1}{\sqrt{x_{0}} }[/tex]

[tex]f'(x) = \dfrac{1}{\sqrt{x} }[/tex]

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