Question

Вычислить производную( срочно)

Answer 1

Ответ:

$y' = ( {ln}^{5} (3x))' \times tg {(1 - 3x)}^{2} + (tg {(1 - 3x)}^{2} )' \times {ln}^{5} (3x) = \\ = 5 {ln}^{4} (3x) \times (ln(3x))' \times (3x)' \times tg {(1 - 3x)}^{2} + \frac{1}{ { \cos }^{2} (1 - 3x) ^{2} } \times ( {(1 - 3x)}^{2} )' \times (1 - 3x) '\times {ln}^{5} (3 x) = \\ = {ln}^{4} (3x) \times (5 \times \frac{1}{3x} \times 3 \times tg {(1 - 3x)}^{2} + \frac{2(1 - 3x) \times ( - 3)}{ { \cos}^{2} {(1 - 3x)}^{2} } \times ln(3x)) = \\ = {ln}^{4} (3x) \times ( \frac{5}{x} tg {(1 - 3x)}^{2} - \frac{6(1 - 3x) \times ln(3x)}{ { \cos }^{2} {(1 - 3x)}^{2} } )$