Question

Задание 5.3.........

Answer 1

Ответ:

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

$b) {(x + y)}^{2} + {(x - 3y)}^{2} = 0 \\ 2(x + y) \times (1 + y') + 2(x - 3y) \times (1 - 3y') = 0 \\ 2x + 2yy' + 2y + 2xy' + 2x + 6xy' - 6y + 18yy'= 0 \\ 4x + 20yy'- 4y + 8xy' = 0 \\ y'(20y + 8x) = - 4x + 4y \\ y' = \frac{4(y - x)}{4(2x + 5y)} \\ y' = \frac{y - x}{2x + 5y}$