Question

Срочно пожалуйста
желательно прописать каждый шаг решения

Answer 1

Ответ:

а)

$y = 4 {x}^{5} - \sqrt[5]{ {x}^{3} } + 2arccosx - 8$

$y' = 4 \times 5 {x}^{4} - \frac{3}{5} {x}^{ - \frac{2}{5} } - \frac{2}{ \sqrt{1 - {x}^{2} } } + 0 = \\ = 20 {x}^{4} - \frac{3}{5 \sqrt[5]{ {x}^{2} } } - \frac{2}{ \sqrt{1 - {x}^{2} } }$

б)

$y= arctgx \times \cos(x) - \frac{ {e}^{x} }{ \sin(x) }$

$y' = (arctgx) '\cos(x) + ( \cos(x))' arctgx - \frac{( {e}^{x} ) ' \sin(x) - ( \sin(x))' \times {e}^{x} }{ { \sin}^{2}(x) } = \\ = \frac{1}{1 + {x}^{2} } \times \cos(x) - \sin(x) \times arctgx - \frac{ {e}^{x} \sin(x) - \cos(x) {e}^{x} }{ { \sin }^{2}(x) } = \\ = \frac{ \cos(x) }{1 + {x}^{2} } - \sin(x) \times arctgx - \frac{ {e}^{x} ( \sin(x) - \cos(x)) }{ { \sin}^{2}(x) }$

в)

$y = tg( ln(6x + 1))$

$y' = \frac{1}{ { \cos}^{2}( ln(6x + 1)) } \times ( ln(6x + 1))' \times (6x + 1)' = \\ = \frac{1}{ { \cos ^{2} ( ln(6x + 1)) } } \times \frac{1}{6x + 1} \times 6 = \\ = \frac{6}{(6x + 1) { \cos}^{2} ( ln(6x + 1)) }$