Question

Знайдіть похідну d^2y/dx^2 функції , заданої параметрично:
x = ctg^2 e^t , y=1/sin e^t

Answer 1

$x = {ctg}^{2} ( {e}^{t} ) \\ y = \frac{1}{ \sin( {e}^{t} ) }$

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

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

$x't = 2ctg( {e}^{t} ) \times ( - \frac{1}{ { \sin}^{2} {e}^{t} } ) \times {e}^{t}$

$y'x = \frac{ {e}^{t} \cos( {e}^{t} ) }{ { \sin }^{2}( {e}^{t} ) } \times ( - \frac{ { \sin}^{2}( {e}^{t} ) }{2ctg( {e}^{t}) {e}^{t} } ) = \\ = - \cos( {e}^{t} ) \times \frac{1}{2} \frac{ \sin( {e}^{t} ) }{ \cos( {e}^{t} ) } = \\ = - \frac{1}{2} \sin( {e}^{t} )$

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

$y''x = - \frac{1}{2} \cos( {e}^{t} ) {e}^{t} \times ( - \frac{ { \sin }^{2} ({e}^{t} ) }{2ctg( {e}^{t} ){e}^{t} } ) = \\ = \frac{1}{4} \cos( {e}^{t} ) \times { \sin }^{2}( {e}^{t} ) \times \frac{ \sin( {e}^{t} ) }{ \cos( {e}^{t} ) } = \\ = \frac{1}{4} { \sin }^{3} ( {e}^{t} )$