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Kovacslore
a fost răspuns

f(x)=x√x_3x...sa se verifice ca f'(x)=3√x-6 supra 2

Răspuns :


[tex]\it f(x) = x\sqrt{x} -3x \\\;\\ \\\;\\ f'(x) = (x\sqrt{x} )' - (3x)' = \sqrt{x} +\dfrac{x}{2\sqrt{x}} -3 = \\\;\\ \\\;\\ = \sqrt{x} + \dfrac{ \sqrt{x} \cdot \sqrt{x}}{2\sqrt{x}} - 3 = \sqrt{x} + \dfrac{ \sqrt{x}}{2} - 3 = \dfrac{2\sqrt{x}+\sqrt{x}-6}{2} =[/tex]


[tex]\it = \dfrac{3\sqrt{x}-6}{2}[/tex]


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