Найти дифференциал тригонометрической функции
Задача. Найти дифференциал \(dy\).
\[y = tg(2\arccos \sqrt {1 - 2{x^2}} ),x > 0.\]
Решение.
\[\begin{array}{l}dy = \left( {\frac{1}{{{{\cos }^2}(2\arccos \sqrt {1 - {x^2}} }} \cdot \left( { - \frac{2}{{\sqrt {1 - (1 - 2{x^2}} }} \cdot \frac{{ - 4x}}{{2\sqrt {1 - 2{x^2}} }}} \right)} \right)dx = \\ = \frac{1}{{{{\cos }^2}(2\arccos \sqrt {1 - 2{x^2}} )}} \cdot \frac{{4x}}{{\sqrt {2{x^2}} \cdot \sqrt {1 - 2{x^2}} }}dx = \\ = \frac{4}{{\sqrt 2 {{\cos }^2}(2\arccos \sqrt {1 - 2{x^2}} ) \cdot \sqrt {1 - 2{x^2}} }}dx.\end{array}\]
\[y = tg(2\arccos \sqrt {1 - 2{x^2}} ),x > 0.\]
Решение.
\[\begin{array}{l}dy = \left( {\frac{1}{{{{\cos }^2}(2\arccos \sqrt {1 - {x^2}} }} \cdot \left( { - \frac{2}{{\sqrt {1 - (1 - 2{x^2}} }} \cdot \frac{{ - 4x}}{{2\sqrt {1 - 2{x^2}} }}} \right)} \right)dx = \\ = \frac{1}{{{{\cos }^2}(2\arccos \sqrt {1 - 2{x^2}} )}} \cdot \frac{{4x}}{{\sqrt {2{x^2}} \cdot \sqrt {1 - 2{x^2}} }}dx = \\ = \frac{4}{{\sqrt 2 {{\cos }^2}(2\arccos \sqrt {1 - 2{x^2}} ) \cdot \sqrt {1 - 2{x^2}} }}dx.\end{array}\]
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