Question

Решите производные ниже

Answer 1

Ответ:

1.

$y '= - 7 {(1 + 2x - {x}^{3}) }^{6} \times (1 + 2x - {x}^{3} )' = \\ = - 7 {(1 + x - {x}^{3} )}^{6} \times (2 - 3 {x}^{2} )$

2.

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

3.

$y' = \frac{1}{ \sqrt{1 - {x}^{2} } } \times (lg(2 {x}^{2} - 1)) + \frac{1}{ ln(10) \times (2 {x}^{2} - 1)} \times 4x \times arcsinx = \\ = \frac{lg(2 {x}^{2} - 1) }{ \sqrt{1 - {x}^{2} } } + \frac{4x \times arcsinx}{ ln(10) \times (2 {x}^{2} - 1) }$

4.

$y' = \frac{1}{ { \cos }^{2}(3x) } \times 3 - \frac{1}{1 - 4x} \times ( - 4) = \\ = \frac{3}{ { \cos }^{2} (3x)} + \frac{4}{1 - 4x}$

5.

$y '= 2(1 + \cos(2x)) \times ( - \sin(2x) ) \times 2 = \\ = - 2 \sin(2x) (1 + \cos(2x))$