Question

с решением тому кто ответит!

Answer 1

1.

$y' = 3 {e}^{3x} \cos(3x) - 3 \sin(3x) {e}^{3x} = 3{e}^{3x} (\cos(3x) - \sin(3x)) \\ y'' = 9 {e}^{3x} ( \cos(3x) - \sin(3x) ) + {e}^{3x} ( - 3 \sin(3x) - 3 \cos(3x) ) = \\ = 3 {e}^{3x} (3 \cos(3x) - 3 \sin(3x) - \sin(3x) - \cos(3x) ) = \\ = 3 {e}^{3x} (2 \cos(3x) - 4 \sin(3x))$

2.

${x}^{2} + \sin(y) = 1 \\ 2x + y \cos(y) = 0 \\ y' \cos(y) = - 2x \\ y' = - \frac{2x}{ \cos(y) } \\$

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

3.

$x't = 2t + 1 \\ y't = \frac{1}{2 \sqrt{t} } \\ \\ y'x = \frac{1}{2 \sqrt{t} } \times \frac{1}{(2t + 1) } = \frac{1}{4t \sqrt{t} + 2 \sqrt{t} }$

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

$(y'x)'t = - {(4 t\sqrt{t} + 2 \sqrt{t}) }^{ - 2} \times (4 \times \frac{3}{ 2 } \sqrt{t} - \frac{1}{ \sqrt{t} } ) = \\ = \frac{6 \sqrt{t} - \frac{1}{ \sqrt{t} } }{ {(4 t\sqrt{t} + 2 \sqrt{t} )}^{2} } = \frac{6t - 1}{ \sqrt{t} {(4 t\sqrt{t} + 2 \sqrt{t} )}^{2} } \\ \\ x't = 2t + 1$

$y''x = \frac{6t - 1}{ \sqrt{t}(2t + 1) {(4 t\sqrt{t} + 2 \sqrt{t} }^{2} ) }$

2 задание

$z = arctg(xy)$

$Z'x = \frac{1}{1 + {x}^{2} {y}^{2} } \times y \\ Z'y = \frac{1}{1 + {x}^{2} {y}^{2} } \times x$

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

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

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