Question

1. Продифференцировать данные функции:
Нужно для зачета, выручите пожалуйста с:

Answer 1

1.

$y = 8 {x}^{2} + \sqrt[3]{ {x}^{4} } - \frac{4}{x} - \frac{2}{ {x}^{3} }$

$y' = 16x + \frac{4}{3} {x}^{ \frac{1}{3} } + 4 {x}^{ - 2} + 6 {x}^{ - 4} = \\ = 16x + \frac{4}{3} \sqrt[3]{x} + \frac{4}{ {x}^{2} } + \frac{6}{ {x}^{4} }$

2.

$y = \frac{3}{ {(x - 4)}^{7} } - \sqrt{5 {x}^{2} - 4x + 3}$

$y' = 3 \times ( - 7) {(x - 4)}^{ - 8} - \frac{1}{2} {(5 {x}^{2} - 4x + 3)}^{ - \frac{1}{2} } \times (10x - 4) = \\ = - \frac{21}{ {(x - 4)}^{8} } - \frac{2(5x - 2)}{2 \sqrt{5 {x}^{2} - 4x + 3 } } = \\ = - \frac{21}{ {(x - 4)}^{8} } - \frac{5x - 2}{ \sqrt{5 {x}^{2} - 4x + 3} }$

3.

$y = {2}^{ \cos(x) } \times arcctg(5 {x}^{3} )$

$y' = ln(2) \times {2}^{ \cos(x) } \times ( - \sin(x) ) \times arcctg(5 {x}^{3} ) + \\ + ( - \frac{1}{1 + 25 {x}^{6} } ) \times 15 {x}^{2} \times {2}^{ \cos(x) } = \\ = - {2}^{ \cos(x) } ( ln(2) \times \sin(x) \times arcctg(5 {x}^{3} ) + \frac{15 {x}^{2} }{25 {x}^{6} + 1 } )$

4.

$y = {e}^{ - 2x} \times {arcsin}^{2} (5x)$

$y' = 2 {e}^{ - 2x} \times {arcsin}^{2} (5x) + \\ + 2arcsin(5x) \times \frac{1}{ \sqrt{1 - 25 {x}^{2} } } \times 5 \times {e}^{ - 2x} = \\ = 2 {e}^{ - 2x} ( {arcsin}^{2} (5x) + \frac{5arcsin(5x)}{ \sqrt{1 - 25 {x}^{2} } }$

5.

$y = arccos( {x}^{2} ) \times ctg(7 {x}^{3} )$

$y '= - \frac{1}{ \sqrt{1 - {x}^{4} } } \times 2x \times ctg(7 {x}^{3} ) - \\ - \frac{1}{ { \sin}^{2} (7 {x}^{3}) } \times 21 {x}^{2} \times arccos( {x}^{2} ) = \\ = - \frac{2xctg(7 {x}^{3}) }{ \sqrt{1 - {x}^{4} } } - \frac{21 {x}^{2}arccos( {x}^{2} ) }{ { \sin}^{2} (7 {x}^{3}) }$