Question

Продифференцировать функции​

Answer 1

Ответ:

$y' = U'V + V'U = \\ = 4 {th}^{3} x \times \frac{1}{ {ch}^{2}(x) } arcctg( \frac{1}{x} ) + ( - \frac{1}{1 + \frac{1}{ {x}^{2} } } ) \times ( - {x}^{ - 2} ) {th}^{4} x = \\ = {th}^{3} (x)( \frac{4arcctg( \frac{1}{x} )}{ {ch}^{2} (x)} - \frac{th(x)}{ {x}^{2}(1 + \frac{1}{ {x}^{2} } ) } ) = \\ = {th}^{3} (x) \times ( \frac{4arcctg( \frac{1}{x}) }{ {ch}^{2} (x)} - \frac{th(x)}{ {x}^{2} + 1} )$