Question

Помогите решить производные
f(x)=(2/x)-(8/√x)+(6/3√x^2)+2x+6x^2√x
Остальное не фото
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Answer 1

$1)f'(x) = \frac{ - 1}{ {x}^{2} } + \frac{4}{ \sqrt{( {x)}^{3} } } - \frac{2}{ \sqrt[3]{x} } + 2 + 9 \sqrt{x} \\ f'(1) = \frac{ - 1}{ {1}^{2} } + \frac{4}{ \sqrt{( {1)}^{3} } } - \frac{2}{ \sqrt[3]{1} } + 2 + 9 \sqrt{1} = \\ = - 1 + 4 - 2 + 2 + 9 = 12$

$2)f'(x) = 2x \sqrt{ {x}^{2} + 1 } + ( {x}^{2} - 2) \frac{x}{\sqrt{ {x}^{2} + 1} } \\ f'( \sqrt{3} ) = 2\sqrt{3} \sqrt{ {\sqrt{3}}^{2} + 1 } + ( {\sqrt{3}}^{2} - 2) \frac{\sqrt{3}}{\sqrt{ {\sqrt{3}}^{2} + 1} } = \\ = 4 \sqrt{3} + \frac{ \sqrt{3} }{2} = \frac{7 \sqrt{3} }{2}$

$3)f'(x) = \frac{9 \sqrt{ {x}^{2} + 1 } - 9x \times x \div \sqrt{ {x}^{2} + 1} }{ {x}^{2} + 1} = \frac{9 \sqrt{ {x}^{2} + 1 } - 9 {x}^{2} \div \sqrt{ {x}^{2} + 1} }{ {x}^{2} + 1} \\f'(2 \sqrt{2} ) = \frac{9 \sqrt{ {(2 \sqrt{2})}^{2} + 1 } - 9 \times {(2 \sqrt{2})}^{2} \div \sqrt{ {(2 \sqrt{2})}^{2} + 1} }{ {(2 \sqrt{2})}^{2} + 1} = \\ = \frac{9 \times 3 - 9 \times 8 \div 3}{9} = 3 - 8 \div 3 = \frac{1}{3}$

Answer 2

Ответ: во вложении Объяснение: