Question

помогите срочно!!До 17:00 60 БАЛОВ​

Answer 1

Ответ:

1

$y' = \cos(5x) \times (5x) '= 5 \cos(5x)$

2

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

3

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

4

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

5

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

6

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

7

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

8

а)

$y '= 6 {(5x - 3)}^{5} \times (5x - 3) '= 30 {(5x - 3)}^{5}$

$y'(1) = 30 \times {2}^{5} = 30 \times 32 = 640$

б)

$y '= \frac{1}{2 \sqrt{10x + 6} } \times 10 = \frac{5}{ \sqrt{10x + 6} }$

$y'(3) = \frac{5}{ \sqrt{30 + 6} } = \frac{5}{6}$