Question

Продифференцировать функцию

Answer 1

Ответ:

$y' = ( {\cth}^{2} (x + 1))' \times \arccos \frac{1}{x} + ( \arccos \frac{1}{x} ) '\times {\cth}^{2} (x + 1) = \\ = 2\cth(x + 1) \times ( - \frac{1}{ {\sh}^{2} (x + 1)} ) \times \arccos \frac{1}{ x} - \frac{1}{ \sqrt{1 - \frac{1}{ {x}^{2} } } } \times ( - {x}^{ - 2} ) \times {\cth}^{2} (x + 1) = \\ = \cth(x + 1) \times ( - \frac{2\arccos \frac{1}{x} }{ {\sh}^{2} (x + 1)} + \sqrt{ \frac{ {x}^{2} }{ {x}^{2} - 1} } \times \frac{1}{ {x}^{2} } \times \cth(x + 1)) = \\ = \cth(x + 1)( \frac{\cth(x + 1)}{x \sqrt{ {x}^{2} - 1 } } - \frac{2\arccos \frac{1}{x} }{ {\sh}^{2}(x + 1) } )$