Найти производную параметрической функции
Задача. Найти производную \({y'_x}\).\[\left\{ \begin{array}{l}x = \sqrt {1 - {t^2}} ,\\y = tg\sqrt {1 + t} .\end{array} \right.\]
Решение.\[\begin{array}{l}{{y'}_x} = \frac{{{{y'}_t}}}{{{{x'}_t}}}.\\x'(t) = \frac{{ - t}}{{\sqrt {1 - {t^2}} }},\\y(t) = \frac{1}{{{{\cos }^2}\sqrt {1 + t} }} \cdot \frac{1}{{2\sqrt {1 + t} }} = \frac{1}{{2\sqrt {1 + t} {{\cos }^2}\sqrt {1 + t} }}.\end{array}\]
\[{y'_x} = - \frac{{\sqrt {1 - {t^2}} }}{{2t\sqrt {1 + t} {{\cos }^2}\sqrt {1 + t} }}.\]
Решение.\[\begin{array}{l}{{y'}_x} = \frac{{{{y'}_t}}}{{{{x'}_t}}}.\\x'(t) = \frac{{ - t}}{{\sqrt {1 - {t^2}} }},\\y(t) = \frac{1}{{{{\cos }^2}\sqrt {1 + t} }} \cdot \frac{1}{{2\sqrt {1 + t} }} = \frac{1}{{2\sqrt {1 + t} {{\cos }^2}\sqrt {1 + t} }}.\end{array}\]
\[{y'_x} = - \frac{{\sqrt {1 - {t^2}} }}{{2t\sqrt {1 + t} {{\cos }^2}\sqrt {1 + t} }}.\]
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