Question

Найти первую и вторую производные

Answer 1

Ответ:

а)

$y' = \frac{7}{5} {x}^{6} - 9 {x}^{ - 4} + 0 = \\ = \frac{7}{5} {x}^{6} - \frac{9}{ {x}^{4} }$

$y ''= \frac{42}{5} {x}^{5} + 36 {x}^{ - 5} = \frac{42}{5} {x}^{5} + \frac{36}{ {x}^{5} }$

б)

$y '= ( - 2 + 2x)tg(6x) + \frac{1}{ { \cos}^{2} 6x} \times 6 (1 - 2x + {x}^{2} ) = \\ = (2x - 2)tg(6x) + \frac{(6 - 12x + 6 {x}^{2}) }{ { \cos}^{2}6x }$

$y ''= 2tg(6x) + \frac{(2 - 2x)}{ { \cos }^{2} 6x} \times 6 + \\ + \frac{( - 12 + 12 {x}^{2}) { \cos }^{2}(6x) - 2 \cos(6x) \times ( - \sin(6x)) \times 6(6 - 12 x + 6 {x}^{2} )}{ { \cos }^{4}6x } = \\ = 2tg(6x) + \frac{12 - 12x}{ { \cos }^{2} (6x)} + \frac{(12 {x}^{2} - 12) { \cos}^{2} (6x) + 6\sin(12x)(6 - 12 x + 6 {x}^{2} ) }{ { \cos}^{4}(6x) } = \\ = 2tg(6x) + \frac{12 - 12x}{ { \cos }^{2}(6x) } + \frac{12 {x}^{2} - 12}{ { \cos }^{2}(6x) } + \frac{6(6 - 12x + 6 {x}^{2}) \sin(12x) }{ { \cos}^{4} (6x)} = \\ = 2tg(6x) + \frac{12 {x}^{2} - 12x}{ { \cos}^{2} (6x)} + \frac{6(6 - 12x + 6 {x}^{2}) \sin(12x) }{ { \cos}^{4}(6x) }$

в)

$y '= \frac{ \cos(2 - x) \times ( - 1) {e}^{3x} - 3{e}^{3x} \sin(2 - x) }{ {e}^{6x} } = \\ = \frac{ - \cos(2 - x) - 3 \sin(2 - x) }{ {e}^{3x} }$

$y'' = \frac{ (\sin(2 - x) \times ( - 1) - 3 \cos(2 - x) \times ( - 1)) {e}^{3x} - 3 {e}^{3x}( - \cos(2 - x) - 3 \sin(2 - x))) }{ {e}^{6x} } = \\ = \frac{ - \sin(2 - x) + 3 \cos(2 - x) + 3 \cos(2 - x) + 9 \sin(2 - x) }{ {e}^{3x} } = \\ = \frac{6 \cos(2 - x) + 8 \sin(2 - x) }{? {e}^{3x} }$

г)

$y' = 2ctg(2x) \times ( - \frac{1}{ { \sin }^{2} (2x)} ) \times 2 = \\ = - \frac{4ctg(2x)}{ { \sin }^{2} (2x)}$

$y ''= - 1 \times \frac{ - \frac{4}{ { \sin }^{2}(2x) } \times 2 \times { \sin }^{2}(2x) - 2 \sin(2x) \cos(2x) \times 2 \times 4ctg(2x) }{ { \sin }^{4} (2x)} = \\ = \frac{8 + 16 { \cos}^{2} (2x)}{ { \sin }^{4} (2x)}$