Question

Записати похідну.Зробіть будь ласка​

Answer 1

Ответ:

1

$y' = (3 {x}^{4} - {x}^{ - 1} + 6 {x}^{ \frac{1}{2} } + \cos(x)) '= \\ = 12 {x}^{3} + {x}^{ - 2} + 6 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - \sin(x) = \\ = 12 {x}^{3} + \frac{1}{ {x}^{2} } + \frac{3}{ \sqrt{x} } - \sin(x)$

2

$y '= (4 {x}^{3} - 2x)'(x + 2) + (x + 2)'(4 {x}^{3} - 2x) = \\ = (12 {x}^{2} - 2)(x + 2) +1 \times (4 {x}^{3} - 2x) = \\ = 12 {x}^{3} + 24 {x}^{2} - 2x - 4 + 4 {x}^{3} - 2x = \\ = 16 {x}^{3} + 24 {x}^{2} - 4x - 4$

3

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

4

$y' = (8 {x}^{ \frac{1}{2} } - 9 {x}^{ - 4} + {x}^{ \frac{1}{3} } )' = \\ = 8 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 9 \times ( - 4) {x}^{ - 5} + \frac{1}{3} {x}^{ - \frac{2}{3} } = \\ = \frac{4}{ \sqrt{x} } + \frac{36}{ {x}^{5} } + \frac{1}{3 \sqrt[3]{ {x}^{2} } }$