Математика: с производными, 1 и 2 задание,

с производными, 1 и 2 задание,

Ответ:

гуфиn

24.04.2021 20:58

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

$y '= ( {5}^{4x} ) '\times { \cos }^{3} (7x - 1) + ( \cos {}^{3} (7x - 1) )' \times {5}^{4x} = \\ = ln(5) \times {5}^{4x} \times 4 \times \cos {}^{3} (7x - 1) + 3 \cos {}^{2} (7 x - 1) \times ( - \sin(7x - 1)) \times 7 \times {5}^{4x} = \\ = {5}^{4x} \cos {}^{2} (7x - 1) \times (4 ln(5) \times \cos(7x - 1) - 21 \sin(7x - 1))$

$y '= 6 {( {tg}^{2} (3x) - 4)}^{5} \times ( {tg}^{2} (3x) - 4)' = \\ = 6 {( {tg}^{2}(3x) - 4 )}^{5} \times (2tg(3x) \times \frac{3}{ \cos {}^{2} (3x) } ) = \\ = \frac{18tg( 3 x)}{ \cos {}^{2} (3x) } {( {tg}^{2} (3x) - 4)}^{5}$

$y = { (\cos(x)) }^{ \sin(x) }$

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

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

$y' = {( \cos(x) )}^{ \sin(x) } \times ( \cos(x) ln( \cos(x) ) + tg(x) \sin(x) ) \\$

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

$y'_x = \frac{y'_t}{x'_t} \\ \\ y'_t = 1 - \frac{1}{ \sin(t) } \times \cos(t) = 1 - ctg(t) \\ x'_t = 1 + \frac{1}{ \cos(t) } \times ( - \sin(t)) = 1 - tg(t)$

$y'_x = \frac{1 - ctgt}{1 - tgt} = \frac{1 - \frac{ \cos(t) }{ \sin(t) } }{1 - \frac{ \sin(t )}{ \cos(t) } } = \\ = \frac{ \sin(t) - \cos(t) }{ \sin(t) } \times \frac{ \cos(t) }{ \cos(t) - \sin(t) } = \\ = - \frac{ \cos(t) }{ \sin(t) } = - ctg(t)$

$y''_{xx} = \frac{(y'_x)'_t}{x'_t} \\$

$(y'_x)'_t = - ( - \frac{1}{ \sin {}^{2} (t) } ) = \frac{1}{ \sin {}^{2} (t) } \\$

$y''_{xx} = \frac{1}{ \sin {}^{2} (t) } \times \frac{1}{1 - tg(t)} = \\ = \frac{1}{ \sin(t) } \times \frac{ \cos(t) }{ \cos(t) - \sin(t) } = \\ = \frac{ctg(t)}{ \cos(t) - \sin(t) }$