Question

Помогите с матешей,найти производную

Answer 1

Ответ:

1.

$y '= 4 {x}^{3} - 6 {x}^{2} + x - 7$

2.

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

3.

$y' = ( {(x - 3)}^{2} )' \times {(x + 1)}^{3} + ( {(x + 1)}^{3} ) '\times {(x - 3)}^{2} = \\ = 2(x - 3) {(x + 1)}^{3} + 3 {(x + 1)}^{2} {(x - 3)}^{2} = \\ = (x - 3) {(x + 1)}^{2} (2(x + 1) + 3(x - 3)) = \\ = (x - 3) {(x + 1)}^{2} (2x + 2 + 3x - 9) = \\ = (x - 3) {(x + 1)}^{2} (5x - 7)$

4.

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

5.

$y' = ( \sin(2x) \cos(2x) )' = (\frac{1}{2} \times 2 \sin(2x) \cos(2x))' = \\ = ( \frac{1}{2} \sin(4x))' = \frac{1}{2} \cos(4x) \times 4 = 2 \cos(4x)$

6.

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

7.

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

8.

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

9.

$y '= 4 {e}^{ {x}^{2} - x} \times (2x - 1)$