Question

Помоги пожалуйста решить =(

Answer 1

Ответ:

1.

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

2.

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

3.

$y = {( {x}^{2} + 1) }^{arctgx}$

формула:

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

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

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

4.

$xy - \cos( {y}^{2} ) + \frac{x}{y} = 0 \\ y + y'x + \sin( {y}^{2} ) \times 2yy' + \frac{y - y'x}{ {y}^{2} } = 0 \\ y.x + 2yy' \sin( {y}^{2} ) - \frac{y'x}{ {y}^{2} } = - y - \frac{1}{y} \\ y'(x + 2y \sin( {y}^{2} ) - \frac{x}{ {y}^{2} } ) = - \frac{ {y}^{2} + 1 }{y} \\ y'\times \frac{x {y}^{2} + 2 {y}^{3} \sin( {y}^{2} ) - x }{ {y}^{2} } = - \frac{ {y}^{2} + 1}{y} \\ y' = - \frac{ {y}^{2} + 1 }{y} \times \frac{ {y}^{2} }{x( {y}^{2} - 1) + 2 {y}^{3} \sin( {y}^{2} ) } \\ y' = - \frac{y( {y}^{2} + 1)}{x( {y}^{2} - 1) + 2 {y}^{3} \sin( {y}^{2} ) }$