Question

Найти производную третьего порядка

$y=(8x+3)\sqrt[8]{8x+3}$

Answer 1

Пошаговое объяснение:

$y = (8x + 3) \sqrt[8]{8x + 3} = {(8x + 3)}^{1 + \frac{1}{8} } = \\ = {(8x + 3)}^{ \frac{9}{8} }$

$y' = \frac{9}{8} {(8x + 3)}^{ \frac{1}{8} } \times 8 = \\ = 9 {(8x + 3)}^{ \frac{1}{8} }$

$y''= 9 \times \frac{1}{8} {(8x + 3)}^{ - \frac{7}{8} } \times 8 = \\ = 9 {(8x + 3)}^{ - \frac{7}{8} }$

$y''' = 9 \times ( - \frac{7}{8} ) {(8x + 3)}^{ - \frac{15}{8} } \times 8 = \\ = - \frac{63}{ \sqrt[8]{ {(8x + 3)}^{15} } }$