Question

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Прошу помощи От умных людей ​

Answer 1

Ответ:

$1)y' = 35 {x}^{4} - \frac{2}{3} {x}^{ - \frac{2}{3} } = 35 {x}^{4} - \frac{2}{3 \sqrt[3]{ {x}^{2} } }$

$2)y' = \frac{1}{3 \sqrt[3]{ {x}^{2} } } \cos(x) - \sqrt[3]{x} \sin(x)$

$3)y' = \frac{ \cos(x)( \cos(x) + 1) + \sin(x) \times \sin(x) }{ {(1 + \cos(x) )}^{2} } = \frac{ { \cos(x) }^{2} + \cos(x) + { \sin(x) }^{2} }{ {(1 + \cos(x)) }^{2} } = \frac{1 + \cos(x) }{ {(1 + \cos(x)) }^{2} } = \frac{1}{1 + \cos(x) }$

$4)y' = - 2 {( \cos(2x)) }^{ - 3} \times ( - \sin(2x)) \times 2 = \frac{4 \sin(2x) }{ { \cos(2x) }^{3} }$

$5)y' = 2(arctg(x) + x) \times ( \frac{1}{1 + {x}^{2} } + 1)$

$6)y' = \frac{1}{ { \cos(5x) }^{2} } \times 5 \sin(7x) + 7 \cos(7x) tg(5x)$

$7)y'=2 \cos(2x) \times ( - \sin(2x)) { \sin(3x) }^{2} + 2 \sin(3x) \cos(3x) { \cos(2x) }^{2} = - \sin(4x) { \sin(3x) }^{2} + \sin(6x) { \cos(2x) }^{2}$

$8)y' = 2(1 + arcsin(x)) \times \frac{1}{ \sqrt{1 - {x}^{2} } }$

$9)y' = \frac{( ln(10) {10}^{x} - ln(10) {10}^{ - x} )2x - 2( {10}^{x} + {10}^{ - x}) }{4 {x}^{2} } = \frac{ ln(10)( {10}^{x} - {10}^{ - x} )x - {10}^{x} - {10}^{ - x} }{2 {x}^{2} }$

$10)y' = \frac{1}{ {x}^{2} - 4x } (2x - 4)$

$11)y' = - \frac{1}{ { \sin( ln(2x) ) }^{2} } \times \frac{1}{2 x } \times 2 = - \frac{1}{x { \sin( ln(2x) ) }^{2} }$