Question

Помогите решить срочно, прошу!!!!!! Дам 50 баллов
Вариант 1

Answer 1

Ответ:

1.

$z = \frac{ ln(x + {y}^{3} ) }{ \sqrt{ {x}^{2} - 1} }$

$z'_x = \frac{( ln(x + {y}^{3} ) )'_x \sqrt{ {x}^{2} - 1} - ( \sqrt{ { x}^{2} - 1} )' \times ln(x + {y}^{3} ) }{ {x}^{2} - 1} = \\ = \frac{ \frac{1}{x + {y}^{3} }\times 1 \times \sqrt{ {x}^{2} - 1 } - \frac{2x}{2 \sqrt{ {x}^{2} - 1 } } \times ln(x + {y}^{3} ) }{ {x}^{2} - 1} = \\ = \frac{1}{ {x}^{2} - 1 } ( \frac{ \sqrt{ {x}^{2} - 1 } }{x + {y}^{3} } - \frac{ ln(x + {y}^{3} ) }{ \sqrt{ {x}^{2} - 1} } ) = \\ = \frac{1}{(x + {y}^{3}) \sqrt{x {}^{2} - 1} } - \frac{ ln(x + {y}^{3} ) }{ \sqrt{ {( {x}^{2} - 1) }^{3} } }$

$z'_y = \frac{1}{ \sqrt{ {x}^{2} - 1 } } \times ( ln(x + {y}^{3} ) )'_y = \\ = \frac{1}{ \sqrt{ {x}^{2} - 1 } } \times \frac{1}{x + {y}^{3} } \times (x + {y}^{3} )'_y = \\ = \frac{3 {y}^{2} }{(x + {y}^{3}) \sqrt{ {x}^{2} - 1 } }$

2.

$z = ln( \frac{1 + \sqrt{ \sin(x) } }{ \sqrt{x {y}^{2} } } )$

$z'_x = \frac{1}{ \frac{1 + \sqrt{ \sin(x) } }{ \sqrt{x {y}^{2} } } } \times \frac{(1 + \sqrt{ \sin(x) } )' \times \sqrt{x {y}^{2} } - y \times ( \sqrt{x} )'(1 + \sqrt{ \sin(x) } )}{x {y}^{2} } = \\ = \frac{ \sqrt{x {y}^{2} } }{1 + \sqrt{ \sin(x)} } \times \frac{ \frac{ \cos(x) }{2 \sqrt{ \sin(x) } } \times y \sqrt{x} - \frac{y}{2 \sqrt{x} } (1 + \sqrt{ \sin(x) } ) }{x {y}^{2} } = \\ = \frac{1}{ \sqrt{x {y}^{2} }(1 + \sqrt{ \sin(x) } )} ( \frac{y \sqrt{x} \cos(x) }{2 \sqrt{ \sin(x) } } - \frac{y(1 + \sqrt{ \sin(x) }) }{2 \sqrt{x} } )$

$z'_y = \frac{ \sqrt{x {y}^{2} } }{1 + \sqrt{ \sin(x) } } \times \frac{1 + \sqrt{ \sin(x) } }{ \sqrt{x} } \times ( \sqrt{ {y}^{2} } ) '= \\ = \frac{ \sqrt{x {y}^{2} } }{ \sqrt{x} } \times 1 = \sqrt{ {y}^{2} } = |y|$

3.

$z = \frac{arccos(x {y}^{2}) }{ \sqrt{x + y} }$

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

$z'_y = \frac{ - \frac{1}{ \sqrt{1 - {x}^{2} {y}^{4} } } \times 2xy \times \sqrt{x + y} - \frac{1}{2 \sqrt{x + y} }arccos(x {y}^{2}) }{x + y} = \\ = - \frac{2xy}{ \sqrt{(x + y)(1 - {xy}^{2} )} } - \frac{arccos(x {y}^{2} )}{2 \sqrt{ {(x + y)}^{3} } }$