Question

помогите пожалуйста решить это​

Answer 1

Ответ:

$f(x) = {x}^{2} \sqrt[4]{225 - {x}^{2} } = {x}^{2} {(225 - {x}^{2}) }^{ \frac{1}{4} }$

$f'(x) = ( {x}^{2} )' \times \sqrt[4]{225 - {x}^{2} } + ( {(225 - {x}^{2}) }^{ \frac{1}{4} } ) '\times (225 - {x}^{2} )' \times {x}^{2} = \\ = 2x \sqrt[4]{225 - {x}^{2} } + \frac{1}{4} {(225 - {x}^{2}) }^{ - \frac{3}{4} } \times ( - 2x) \times {x}^{2} = \\ = 2x \sqrt[4]{225 - {x}^{2} } - \frac{ {x}^{3} }{2 \sqrt[4]{ {(225 - {x}^{2}) }^{3} } }$

$f(12) = 2 \times 14 \sqrt[4]{225 - 144} - \frac{ {12}^{2} \times 12}{2 \sqrt[4]{ {(225 - 144)}^{3} } } = \\ = 28 \sqrt[4]{81} - \frac{ {12}^{2} \times 6 }{ \sqrt[4]{ {81}^{3} } } = 28 \times 3 - \frac{ {12}^{2} \times 6 }{ \sqrt[4]{ {( {3}^{4} )}^{3} } } = \\ = 84 - \frac{ {12}^{2} \times 6}{ {3}^{3} } = 84 - \frac{12 \times 12 \times 6}{3 \times 3 \times 3} = \\ = 84 - 32 = 52$