Алгебра. Решите пожалуйста

Question 1

Question 2

6. $z = arctg{ \frac{x}{y} } \ ; \ \ \ x = e^{2t} + t^5 \ , \ \ \ y = e^{2t} - 1 \ ;$

$\frac{dz}{dt} = \frac{d}{dt} arctg{ \frac{x}{y} } = \frac{1}{ 1 + (x/y)^2 } \cdot \frac{d}{dt} \frac{x}{y} = \frac{y^2}{ x^2 + y^2 } \cdot \frac{ x'_t y - y'_t x }{ y^2 } = \\\\ = \frac{ ( e^{2t} + t^5 )'_t y - ( e^{2t} - 1 )'_t x }{ x^2 + y^2 } = \frac{ 2e^{2t} y + 5t^4 y - 2e^{2t} x }{ x^2 + y^2 } = \\\\ = \frac{ 2e^{2t} ( y - x ) + 5t^4 y }{ [ e^{2t} + t^5 ]^2 + [ e^{2t} - 1 ]^2 } = \frac{ 2e^{2t} ( -1 - t^5 ) + 5t^4 ( e^{2t} - 1 ) }{ e^{4t} + 2 e^{2t} t^5 + t^{10} + e^{4t} - 2 e^{2t} + 1 } = \\\\ = \frac{ -2e^{2t} - 2e^{2t} t^5 + 5t^4 e^{2t} - 5t^4 }{ 2e^{4t} + 2 e^{2t} ( t^5 - 1 ) + t^{10} + 1 } = \frac{ 5t^4 ( e^{2t} - 1 ) - 2e^{2t} ( 1 + t^5 ) }{ 2e^{4t} + 2 e^{2t} ( t^5 - 1 ) + t^{10} + 1 } \ ;$

$\frac{dz}{dt} = \frac{ 5t^4 ( e^{2t} - 1 ) - 2e^{2t} ( 1 + t^5 ) }{ 2e^{4t} + 2 e^{2t} ( t^5 - 1 ) + t^{10} + 1 } \ ;$

7. $z = x^2 y^2 \ ; \ \ \ x = u + 2v \ ; \ \ \ y = \frac{v}{u} \ ;$

$z = (xy)^2 = ( ( u + 2v ) \frac{v}{u} )^2 = ( u \cdot \frac{v}{u} + 2v \cdot \frac{v}{u} )^2 = ( v + \frac{ 2v^2 }{u} )^2 \ ;$

$\frac{ \partial z }{ \partial u } = \frac{ \partial }{ \partial u } ( v + \frac{ 2v^2 }{u} )^2 = 2 ( v + \frac{ 2v^2 }{u} ) \cdot \frac{ \partial }{ \partial u } ( v + \frac{ 2v^2 }{u} ) = \\\\ = 2 ( v + \frac{ 2v^2 }{u} ) \cdot ( -\frac{ 2v^2 }{u^2} ) = -4 ( \frac{v}{u} )^2 ( v + \frac{ 2v^2 }{u} ) \ ;$

$\frac{ \partial z }{ \partial v } = \frac{ \partial }{ \partial v } ( v + \frac{ 2v^2 }{u} )^2 = 2 ( v + \frac{ 2v^2 }{u} ) \cdot \frac{ \partial }{ \partial v } ( v + \frac{ 2v^2 }{u} ) = \\\\ = 2 ( v + \frac{ 2v^2 }{u} ) \cdot ( 1 + \frac{ 4v }{u} ) = 2 ( v + \frac{ 2v^2 }{u} ) ( 1 + \frac{ 4v }{u} ) \ ;$