Question

Алгебра 30 баллов. Заранее спасибо!
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Answer 1

Ответ:

Объяснение:

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Answer 2

$1)f(x)=x^{\frac{1}{3} }\\\\f'(x)=(x^{\frac{1}{3} })'=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}=\frac{1}{3\sqrt[3]{x^{2}}}=\frac{\sqrt[3]{x}}{3x}\\\\2)h(x)=5x^{-\frac{2}{5}}\\\\h'(x)=5(x^{-\frac{2}{5} })'=5*(-\frac{2}{5})*x^{-\frac{2}{5}-1 }=-2x^{-\frac{7}{5}}=-\frac{2}{\sqrt[5]{x^{7}}}=-\frac{2}{x\sqrt[5]{x^{2}}}\\\\3)f(x)=x^{\frac{3}{2}}\\\\f'(x)=(x^{\frac{3}{2}})'=\frac{3}{2}x^{\frac{1}{2}}=\frac{3\sqrt{x}}{2}$

$4)f(x)=\sqrt[5]{x^{3}}=x^{\frac{3}{5}}\\\\f'(x)=(x^{\frac{3}{5}})'=\frac{3}{5}x^{\frac{3}{5}-1 }=\frac{3}{5}x^{-\frac{2}{5}}=\frac{3\sqrt[5]{x^{2}}}{5} \\\\5)f(x)=\frac{1}{2x^{\frac{2}{9}}}=\frac{1}{2}x^{-\frac{2}{9}} \\\\f'(x)=\frac{1}{2}(x^{-\frac{2}{9}})'=\frac{1}{2}*(-\frac{2}{9})*x^{-\frac{2}{9}-1 }=-\frac{1}{9}x^{-\frac{11}{9}}=-\frac{1}{9x^{\frac{11}{9}}}=-\frac{1}{9\sqrt[9]{x^{11}}}=-\frac{1}{9x\sqrt[9]{x^{2}}}$

$6)g(x)=\frac{7}{\sqrt[4]{x}}=7x^{-\frac{1}{4}}\\\\g'(x)=7(x^{-\frac{1}{4}})'=7*(-\frac{1}{4})x^{-\frac{1}{4}-1 }=-\frac{7}{4}x^{-\frac{5}{4}}=-\frac{7}{4\sqrt[4]{x^{5}}}=-\frac{7}{x\sqrt[4]{x}}$