Question

знайти похідну функції ​

Answer 1

Ответ:

1

$y' = \frac{1}{2} \times 2x + 3 = x + 3$

2

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

3

$y' = \frac{(3 {x}^{2} + 2)' \times ctgx - (ctgx) '\times (3 {x}^{2} + 2)}{ {ctg}^{2}x } = \\ = \frac{6x \times ctgx + \frac{3 {x}^{2} + 2 }{ { \sin }^{2}x } }{ {ctg}^{2}x } = \frac{6x}{ctgx} + \frac{3 {x}^{2} + 2 }{ { \cos }^{2}x }$

4

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

5

$y' = 3 {(3x - 2)}^{2} \times 3 \times \cos(2x) - 2 \sin(2x) \times {(3x - 2)}^{3} = \\ = {(3x - 2)}^{2} (9 \cos(2x) - 2(3x - 2)\sin(2x) ) = \\ = {(3x - 2)}^{2} (9 \cos(2x) - (6x - 4) \sin(2x))$

6

$y' = \frac{2x \cos( {x}^{2} - 4 ) \times tg2x - \frac{2}{ { \cos}^{2} (2x)} \times \sin( {x}^{2} - 4 ) }{ {tg}^{2}2x } = \\ = \frac{2x \cos( {x}^{2} - 4 ) }{tg2x} - \frac{ \sin( {x}^{2} - 4 ) }{ { \sin}^{2}(2x) }$

7

$y' = \frac{5}{5x - 3} \times {5}^{ \cos(x ) } + ln(5) \times {5}^{ \cos(x) } \times ( - \sin(x)) \times ln(5x - 3) - \frac{5}{ \cos ^{2} (5x) } = \\ = {5}^{ \cos(x) } ( \frac{5}{5x - 3} - ln(5(5x - 3)) \sin(x) ) - \frac{5}{ { \cos }^{2} (5x)}$