Question

Найти производные dy/dx данных функций

Answer 1

а)

$y' = \frac{5}{2} {x}^{ \frac{3}{2} } + 3 {x}^{ - 2} - 12 {x}^{ - 4} - 9 {x}^{2} = 2.5x \sqrt{x} + \frac{3}{ {x}^{2} } - \frac{12}{ {x}^{4} } - 9 {x}^{2}$

б)

$y' = \frac{6}{5} {(x - 2)}^{ \frac{1}{5} } + 3 {(7 {x}^{3} - {x}^{2} - 4)}^{ - 2} \times (21 {x}^{2} - 2x) = \frac{6}{5} \sqrt[5]{x - 2} + \frac{3(21 {x}^{2} - 2x)}{ {(7 {x}^{3} - {x}^{2} - 4)}^{2} }$

в)

$y' = 3 { \sin(7x) }^{2} \times \cos(7x) \times 7 \times arcctg(5 {x}^{2} ) - \frac{1}{1 + 25 {x}^{4} } \times 10x \times { \sin(7x) }^{3} = \\ { \sin(7x) }^{2} (21 \cos(7x) arcctg(5 {x}^{2} ) - \frac{10x \sin(7x) }{1 + 25 {x}^{4} }$

г)

$y' = \frac{ {e}^{ - tg(3x)} \times \frac{( - 1)}{ { \cos(3x) }^{2} } \times 3 \times (4 {x}^{2} - 3x + 5) - (8x - 3) {e}^{ - tg(3x)} }{{(4 {x}^{2} - 3x + 5)}^{2} } = \\ \frac{ {e}^{ - tg(3x)}( \frac{ - 3(4 {x}^{2} - 3x + 5) }{ { \cos(3x) }^{2} } - 8x + 3) }{ {(4 {x}^{2} - 3x + 5)}^{2} } = \frac{ - \frac{3(4 {x}^{2} - 3x + 5)}{ { \cos(3x) }^{2} } - 8x + 3}{ {e}^{ tg(3x)} {(4 {x}^{2} - 8x + 5) }^{2} } = \\ - \frac{3}{ { \cos(3x) }^{2} {e}^{tg(3x)} (4 {x}^{2} - 8x + 5) } - \frac{8x - 3}{ {e}^{tg(3x)} {(4 {x}^{2} - 8x + 5)}^{2} }$