Question

СРОЧНОООООООООО!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Ответ:

$x = arctgt \\ y = arctg \sqrt{t - 1}$

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

$y't = \frac{1}{1 + t - 1} \times \frac{1}{2 \sqrt{t - 1} } = \\ = \frac{1}{2t \sqrt{t - 1} }$

$x't = \frac{1}{1 + {t}^{2} }$

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

при t = 2

$y'x(2) = \frac{1 + 4}{4 \times 1} = \frac{5}{4} = 1.25$