Question

ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО

Answer 1

Ответ:

$1)y' = 4 {( \frac{ {x}^{2} - 1}{ {x}^{2} + 1} )}^{3} \times \frac{2x( {x}^{2} + 1) - 2x( {x}^{2} - 1) }{ {( {x}^{2} + 1)}^{2} } = 4 \frac{ {( {x}^{2} - 1)}^{3} }{ {( {x}^{2} + 1) }^{3} } \times \frac{2x( {x}^{2} + 1 - {x}^{2} + 1) }{ {( {x}^{2} + 1)}^{2} } = \frac{16x {( {x}^{2} - 1) }^{3} }{ {( {x}^{2} + 1)}^{5} }$

$2)y' = 4 {x}^{3} (8 { ln(x) }^{2} - 4 ln(x) + 1) + {x}^{4} (16 ln(x) \times \frac{1}{x} - \frac{4}{ x } ) = \\ = {x}^{3} (32 { ln(x) }^{2} - 16 ln(x) + 4 + 16ln(x) - 4) = \\ = {x}^{3} \times 32 { ln(x) }^{2}$

$3)y' = \frac{1}{3} {x}^{ - \frac{2}{3} } ( {e}^{3x} - 5) + \sqrt[3]{x} \times 3 {e}^{3x} = \\ = \frac{ {e}^{3x} - 5 }{3 \sqrt[3]{ {x}^{2} } } + 3 \sqrt[3]{x} {e}^{3x}$

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

$5)y' = \frac{1}{ \sqrt[3]{ { (\frac{1 - 3x}{1 + 3x}) }^{2} } } \times \frac{2}{3} {( \frac{1 - 3x}{1 + 3x} )}^{ - \frac{1}{3} } \times \frac{ - 3(1 + 3x) - 3(1 - 3x)}{ {(1 + 3x)}^{2} } = \\ = \sqrt[3]{ { (\frac{1 + 3x}{1 - 3x}) }^{2} } \times \frac{2}{3} \times \sqrt[3]{ \frac{1 + 3x}{1 - 3x} } \times \frac{ - 3(1 + 3x + 1 - 3x)}{ {(1 + 3x)}^{2} } = \\ = \frac{2}{3} \times \frac{1 + 3x}{1 - 3x} \times \frac{ - 3 \times 2}{ {(1 + 3x)}^{2} } = \\ = - \frac{4}{(1 - 3x)(1+ 3x)} = - \frac{4}{1 - 9 {x}^{2} }$

$6)y' = \frac{1}{ \frac{x(1 + {x}^{2}) }{ \sqrt{1 - {x}^{2} } } } \times ( \frac{x + {x}^{3} }{ \sqrt{1 - {x}^{2} } } )' = \\ = \frac{ \sqrt{1 - {x}^{2} } }{x(1 + {x}^{2} )} \times \frac{(1 + 3 {x}^{2}) \sqrt{1 - {x}^{2} } + \frac{2x}{2 \sqrt{1 - {x}^{2} } } (x + {x}^{3}) }{1 - {x}^{2} } = \\ = \frac{(1 + 3 {x}^{2}) \sqrt{1 - {x}^{2} } + \frac{ {x}^{2} (1 + {x}^{2}) }{ \sqrt{1 - {x}^{2} } } }{x(1 + {x}^{2} ) \sqrt{1 - {x}^{2} } } = \\ = \frac{(1 + 3 {x}^{2} )}{x(1 + {x}^{2} )} + \frac{x}{1 - {x}^{2} }$

$7)y' = 3 ln(1 - {x}^{2} ) + \frac{3x}{1 - {x}^{2} } \times ( - 2x) = \\ = 3 ln(1 - {x}^{2} ) - \frac{6 {x}^{2} }{1 - {x}^{2} }$