Question

Найти производную срочно, 50 баллов

Answer 1

Ответ:

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