Question

Найти первую производную x y' заданных функций:

Answer 1

Ответ:

1)

$y = \frac{4 + 3 {x}^{2} }{x \times \sqrt[3]{ {(2 + {x}^{3} )}^{2} } }$

С помощью логарифмической производной:

$y' = ( ln(y)) ' \times y$

$( ln(y))' = ( ln( \frac{4 + 3 {x}^{2} }{x \sqrt[3]{ {(2 + {x}^{3} )}^{2} } } ) ' = \\ = ( ln(4 + 3 {x}^{2} ) - ln(x) - ln( {(2 + {x}^{3}) }^{ \frac{2}{3} } ) ) '= \\ = \frac{6x}{4 + 3 {x}^{2} } - \frac{1}{x} - \frac{2 \times 3 {x}^{2} }{3(2 + {x}^{2}) } = \\ = \frac{18{x}^{2}(2 + {x}^{3} ) - 3(2 + {x}^{3})(4 + 3 {x}^{2}) - 6 {x}^{3}(4 + 3 {x}^{2} ) }{3x(4 + 3 {x}^{2})(2 + {x}^{3}) } = \\ = \frac{3( - 3 {x}^{5} - 12 {x}^{3} + 6 {x}^{2} - 8) }{3x(4 + 3 {x}^{2} )(2 + {x}^{3}) } = \\ = - \frac{3 {x}^{5} + 12 {x}^{3} - 6 {x}^{2} + 8 }{x(4 + 3 {x}^{2} )(2 + {x}^{3}) }$

$y = \frac{4 + 3 {x}^{2} }{x \times {(2 + {x}^{3} )}^{ \frac{2}{3} } } \times ( - \frac{3 {x}^{5} + 12 {x}^{3} - 6 {x}^{2} + 8 }{x(4 + 3 {x}^{2})(2 + {x}^{3}) } ) = \\ = - \frac{3 {x}^{5} + 12 {x}^{3} - 6 {x}^{2} + 8}{ {x}^{2} \sqrt[3]{ {(2 + {x}^{3}) }^{5} } }$

б)

$y' = 3 {ln}^{2} (1 + \cos(x)) \times \frac{1}{1 + \cos(x) } ( - \sin(x)) + \frac{1}{6} \times \frac{27 {e}^{x} {( {e}^{x} + 1) }^{3} - 3 {( {e}^{x} + 1)}^{2} \times {e}^{x} \times (27 {e}^{x} + 11) }{ {( {e}^{x} + 1) }^{6} } = \\ = - \frac{ 3 \sin(x) {ln}^{2}(1 + \cos(x)) }{1 + \cos(x) } + \frac{1}{6} \times \frac{ {(1 + {e}^{x}) }^{2} {e}^{x} (27 ( {e}^{x} + 1) - 3(27 {e}^{x} + 11)) }{ {( {e}^{x} + 1) }^{6} } = \\ = - \frac{3 \sin(x) {ln}^{2} (1 + \cos(x)) }{1 + \cos(x) } + \frac{1}{6} \times \frac{ {e}^{x} (27 {e}^{x} + 27 - 81 {e}^{x} - 33 )}{ {( {e}^{x} + 1) }^{4} } = \\ = - \frac{3 \sin(x) {ln}^{2}(1 + \cos(x)) }{1 + \cos(x) } + \frac{ {e}^{x} ( - 54 {e}^{x} - 6)}{6 {( {e}^{x} + 1)}^{4} } = \\ = - \frac{3 \sin(x) {ln}^{2}(1 + \cos(x)) }{1 + \cos(x) } - \frac{ {e}^{ x } (9 {e}^{x} + 1) }{ {( {e}^{x} + 1)}^{2} }$

в)

$y' = - \sin(ctg2) - \frac{1}{16} \times \frac{2 \cos(8x) \times ( - \sin(8x)) \times 8 \times \sin(16x) - 16 \cos(16x) \times { \cos }^{2} (8x)}{ { \sin}^{2}(16x) } = \\ = - \sin(ctg2) - \frac{ - 8 \sin(16x) \times \sin(16x) - 16 \cos(16x) { \cos}^{2}(8x) }{ {16 \sin }^{2}(16x) } = \\ = - \sin(ctg2) + \frac{1}{2} + \frac{ \cos(16x) { \cos }^{2}(8x) }{ { \sin }^{2} (16x)} = \\ = \frac{1}{2} - \sin(ctg2) + \frac{ { \cos}^{2}(8x) ctg(16x)}{ \sin(16x) }$

г)

$y' = \frac{1}{2} ( \sqrt{8x - {x}^{2} -7 } + (x - 4) \times \frac{1}{2} {(8x - {x}^{2} - 7)}^{ - \frac{1}{2} } \times (8 - 2x)) + 9 \times \frac{1}{ \sqrt{1 - \frac{x - 1}{6} } } \times \frac{1}{6} = \\ = \frac{ \sqrt{8x - {x}^{2} - 7 } }{2} + \frac{(x - 4)(4 - x)}{2 \sqrt{8x - {x}^{2} - 7 } } + \sqrt{ \frac{6}{6 - x + 1} } \times \frac{1}{6} = \\ = \frac{ \sqrt{8x - {x}^{2} - 7 } }{2} - \frac{ {(x - 4)}^{2} }{2 \sqrt{8x - {x}^{2} - 7 } } + \frac{1}{ \sqrt{6(7 - x)} }$