Question

Найти производные заданных функций

Answer 1

Ответ:

в

$y' = x'\arccos(2x) +( \arccos(2x)) '\times x - \frac{1}{2} \times \frac{1}{2} {(1 - 4 {x}^{2} )}^{ - \frac{1}{2} } \times ( - 8x) = \\ = \arccos(2x) - \frac{1}{ \sqrt{1 - 4 {x}^{2} } } \times 2 \times x - \frac{1}{4 \sqrt{1 - 4 {x}^{2} } } \times ( - 8x) = \\ = \arccos(2x) - \frac{2x}{ \sqrt{1 - 4 {x}^{2} } } + \frac{2x}{ \sqrt{1 - 4 {x}^{2} } } = \arccos(2x)$

г

$y' = ln(2) \times {2}^{ {ctg}^{2} (3x)} \times 2ctg(3x) \times ( - \frac{1}{ \sin {}^{2} (3x) } ) \times 3 = \\ = - \frac{3ctg(3x)}{ \sin {}^{2} (3x) } \times ln(2) \times {2}^{ {ctg}^{2} (3x)}$

д

$y = {(x + 5)}^{ {x}^{2} - 4x - 5}$

$y '= ( ln(y))' \times y$

$( ln(y)) ' = ( ln(x + 5 ){}^{ {x}^{2} - 4x - 5} ) '= \\ = (( {x}^{2} - 4x - 5) \times ln(x + 5)) ' = \\ = (2x - 4) ln(x + 5) + \frac{1}{x + 5} \times ( {x}^{2} - 4x - 5)$

$y' = {(x + 5)}^{ {x}^{2} - 4x - 5 } \times ((2x - 4) ln(x + 5) + \frac{ {x}^{2} - 4x - 5}{x + 5} )$