Найти вторую производную функции y(x):. Created by marikiri01. algebra-ru

Найти вторую производную функции y(x):...

marikiri01 @marikiri01

July 2022 1 6 Report

Найти вторую производную функции y(x):

Answers & Comments

Miroslava227

Ответ:

а)

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

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

б)

$y'x = \frac{y't}{x't}$

$y''x = \frac{(y'x)'t}{x't}$

$y't = - \sin(t) \\ x't = \frac{1}{ \sin(t) } \times \cos(t)$

$y'x = \frac{ - \sin(t) }{ \frac{ \cos(t) }{ \sin(t) } } = - \frac{ { \sin }^{2} (t)}{ \cos(t) }$

$(y'x)'t = - \frac{2 \sin(t) \cos(t) \times \cos(t) - ( - \sin(t)) \times { \sin }^{2}(t) }{ { \cos }^{2}(t) } = \\ = - \frac{ \sin(t)( { \cos}^{2} (t) + { \sin}^{2} (t))}{ { \cos}^{2} (t)} = - \frac{ \sin(t) }{ { \cos}^{2}(t) }$

$y''x = - \frac{ \sin(t) }{ { \cos}^{2} (t)} \times \frac{ \sin(t) }{ \cos(t) } = - \frac{ { \sin }^{2} (t)}{ { \cos}^{3}(t) }$

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