Question

Помогите, пожалуйста!
Найти производную dy/dx, если:

Answer 1

Ответ:

а)

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

б)

$y' = \frac{1}{6} {( ln( \sin( {3}^{x} {x}^{2} ) ) }^{ - \frac{5}{6} } \times \frac{1}{ \sin( {3}^{x} {x}^{2} ) } \cos( {3}^{x} {x}^{2} ) \times ( ln(3) \times {3}^{x} {x}^{2} + 2x \times {3}^{x} ) = \\ = \frac{ctg( {3}^{x} {x}^{2}) }{6 \sqrt[6]{ {ln}^{5} ( \sin( {3}^{x} {x}^{2} )) } } \times {3}^{x} ( ln(3) {x}^{2} + 2x)$

в)

${x}^{3} y + {y}^{3} x = x - {y}^{2}$

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

г)

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

$y't = \frac{1}{2} \times \frac{1}{ \sqrt{1 - {e}^{2t} } } \times ( - 2 {e}^{2t} ) = \\ = - \frac{ {e}^{2t} }{ \sqrt{1 - {e}^{2t} } }$

$x't = \frac{1}{ \sqrt{1 - {e}^{2t} } } \times {e}^{t}$

$y'x = \frac{ - {e}^{2t} }{ \sqrt{1 - {e}^{2t} } } \times \frac{ \sqrt{1 - {e}^{2t} } }{ {e}^{t} } = \\ = - {e}^{t}$