Question

расписать решение, как можно быстрее пожалуйста ​

Answer 1

Ответ:

1.

$y = log_{3}(6x)$

$y' = \frac{1}{ ln(3) \times (6x)} \times (6x)' = \frac{6}{ ln(3) \times 6x} = \\ = \frac{1}{x ln(3) }$

2.

$y = lg(x) \\ y' = \frac{1}{ ln(10) \times x}$

3.

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

4.

$\\ y' = x' ln(x) + ( ln(x)) ' \times x = \\ = ln(x) + x \times \frac{1}{x} = ln(x) + 1$

5.

$y '= (6 {x}^{ - 5} - 7 {x}^{ - 8} ) '= \\ = - 30 {x}^{ - 6} + 56 {x}^{ - 9} = \\ = - \frac{30}{ {x}^{6} } + \frac{56}{ {x}^{9} }$

6.

$y' = (6 {x}^{ - \frac{5}{6} } )' = 6 \times ( - \frac{5}{6} ) {x}^{ - \frac{11}{6} } = \\ = - \frac{5}{x \sqrt[6]{ {x}^{5} } }$

7.

$y '= (\frac{4}{ {(2x - 8)}^{3} } )' = (4 {(2x - 8)}^{ - 3} ) '= \\ = - 12 {(2x - 8)}^{ - 4} \times 2 = \\ = - \frac{24}{ {(2x - 8)}^{4} }$