Question

Алгебра 10 класс, пожалуйста помогите номер 36.5

Answer 1

Ответ:

1

$y = \sqrt[9]{x} = {x}^{ \frac{1}{9} }$

$y' = \frac{1}{9} {x}^{ \frac{1}{9} - 1} = \frac{1}{9} {x}^{ - \frac{8}{9} } = \frac{1}{9 \sqrt[9]{ {x}^{8} } }$

2

$y = \sqrt[6]{ {x}^{5} } = {x}^{ \frac{5}{6} }$

$y' = \frac{5}{6} {x}^{ \frac{5}{6} - 1} = \frac{5}{6} {x}^{ - \frac{1}{6} } = \frac{5}{6 \sqrt[6]{x} }$

3

$y = \frac{1}{ \sqrt[12]{ {x}^{7} } } = {x}^{ - \frac{7}{12} }$

$y' = - \frac{7}{12} {x}^{ - \frac{7}{12} - 1 } = - \frac{7}{12} {x}^{ - \frac{19}{12} } = \\ = - \frac{7}{12 \sqrt[12]{ {x}^{19} } } = - \frac{7}{12x \sqrt[12]{ {x}^{7} } }$