Question

знайдіть похідну функції даю 50 балов​

Answer 1

Ответ:

1

$y' = 5$

2

$y' = 2x$

3

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

4

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

5

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

6

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

7

$f'(x) = - \sin(5 - 3x) \times (5 - 3x)' = \\ = - \sin(5 - 3x) \times ( - 3) = \\ = 3 \sin(5 - 3x)$

8

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

9

$f'(x) = ( \sin(x) \cos(5x) + \cos(x) \sin(5x) )' = \\ = ( \sin(x + 5x) ) '= (\sin(6x) )' = \\ = 6\cos(6x)$