Question

Помогите пожалуйсто,срочно
вместо N нужно подставить цифру 8
(N=8)

Answer 1

Ответ:

$1)y = \sqrt[8]{8x + 5} \\ y' = \frac{1}{8} {(8x + 5)}^{ - \frac{7}{8} } \times 8 = \frac{1}{ \sqrt[8]{ {(8x + 5)}^{7} } }$

$2)y = {(8 {x}^{2} - 8) }^{8} \\ y' = 8 {(8 {x}^{2} - 8)}^{7} \times 16x = 128x {(8 {x}^{2} - 8)}^{7}$

$3)y = 5 (6x - 8 {x}^{4}) \\ y' = 5(6 - 32 {x}^{3} ) = 30 - 160 {x}^{3}$

$4)y = \frac{8}{ {(8 {x}^{8} - 8x) }^{8} } \\ y' = 8 \times ( - 8) {(8 {x}^{8} - 8x) }^{ - 9} \times (64 {x}^{7} - 8) = \frac{ - 64(64 {x}^{7} - 8)}{ {(8 {x}^{8} - 8x) }^{9} }$

$5)y = \frac{1}{ \sqrt[8]{ {( {x}^{8} + 8 {x}^{2} - 7) }^{5} } } \\ y' = - \frac{5}{8} {( {x}^{8} + 8 {x}^{2} - 7) }^{ - \frac{13}{8} } \times (8 {x}^{7} + 16x) = \frac{ - 5 \times 8( {x}^{7} + 2x) }{8 \sqrt[8]{ {( {x}^{8} + 8 {x}^{2} + 7) }^{13} } } = \frac{ - 5( {x}^{7} + 2x)}{ \sqrt[8]{ {( {x}^{8} + 8 {x}^{2} - 7) }^{13} } }$

$7)y = \sin( \frac{8x + 5}{7 -8x } ) \\ y' = \cos( \frac{8x + 5}{7 - 8x} ) \times ( \frac{8(7 - 8x) + 8(8x + 5)}{ {(7 - 8x)}^{2} }) = \cos( \frac{8x + 5}{7 - 8x} ) \times ( \frac{56 - 64x + 64x + 40}{ {(7 - 8x)}^{2} } = \cos( \frac{8x + 5}{7 - 8x} ) \times \frac{96}{ {(7 - 8x)}^{2} }$