Question

найти производную
45.1;45.2;45.3​ (алгебра)

Answer 1

Ответ:

45.1

1

$f'(x) = \cos(3x) \times (3x) '= 3 \cos(3x)$

2

$f'(x) = - \sin(1 - 2x) \times (1 - 2x)' = 2 \sin(1 - 2x)$

3

$f'( x) = \frac{1}{ \cos {}^{2} (5x) } \times (5x) '= \frac{5}{ \cos {}^{2} (5x) }$

4

$f'(x) = - \frac{1}{ \sin {}^{2} (x - 2) }$

5

$f'(x) = \cos(3 - 2x) \times (3 - 2x) '= - 2 \cos(3 - 2x)$

6

$f'(x) = - \frac{1}{ \sin {}^{2} (5 - 3x) } \times (5 - 3x) '= \frac{3}{ \sin {}^{2} (5 - 3x) }$

45.2

1

$f'(x) = 2(3x - 1) \times (3x - 1) '= \\ = 2(3x - 1) \times3 = 6(3x - 1) = 18x - 6$

2

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

3

$f'(x) = - 3 {(2 - 3x)}^{ - 4} \times (2 - 3x)' = \\ = \frac{9}{ {(2 - 3x)}^{4} }$

4

$f'(x) = 0 - ( - 4) {(1 + 2x)}^{ - 5} \times (1 + 2x)' = \\ = \frac{8}{ {(1 + 2x)}^{5} }$

5

$f'(x) =5 - 2 {(1 - 3x)}^{ - 3} \times (1 - 3x)' = \\ = 5 + \frac{6}{ {(1 - 3x)}^{3} }$

6

$f'(x) = 2x - 2 {(1 + 5x)}^{ - 3} \times (1 + 5x)' = \\ = 2x - \frac{10}{ {(1 + 5x)}^{3} }$

45.3

1

$f'(x) = ( {(2x - 5)}^{ \frac{1}{2} } ) = \\ = \frac{1}{2} {(2x - 5)}^{ - \frac{1}{2} } \times (2x - 5) '= \\ = \frac{1}{2 \sqrt{2x - 5} } \times 2 = \frac{1}{ \sqrt{2x - 5} }$

2

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

3

$f'(x) = \frac{1} {2} {(2 - 5x)}^{ - \frac{1}{2} } \times (2 - 5x)' = \\ = - \frac{5}{2 \sqrt{2 - 5x} }$

4

$f'(x) = \frac{1}{2 \sqrt{3 {x}^{2} - 5x} } \times (3 {x}^{2} - 5x)' = \\ = \frac{6x - 5}{2 \sqrt{3 {x}^{2} - 5x } }$

5

$f'(x) = \frac{1}{2 \sqrt{3 {x}^{2} - 5x + 1 } } \times (6x - 5) = \\ = \frac{6x - 5}{2 \sqrt{3 {x}^{2} - 5x + 1} }$

6

$f'(x) = \frac{1}{2 \sqrt{2 - 3 {x}^{2} + 5x} } \times ( - 6x + 5) = \\ = - \frac{6x - 5}{2 \sqrt{2 - 3 {x}^{2} + 5x} }$