Математика: ДИФФЕРЕНЦИАЛЬНОЕ ИСЧИСЛЕНИЕ ФУНКЦИИ ОДНОЙ ПЕРЕМЕННОЙ

ДИФФЕРЕНЦИАЛЬНОЕ ИСЧИСЛЕНИЕ ФУНКЦИИ ОДНОЙ ПЕРЕМЕННОЙ

Ответ:

Diana102004

23.01.2021 21:12

$y' = \frac{1}{3} {x}^{ - \frac{2}{3} } + 5 \times \frac{3}{4} {x}^{ - \frac{1}{4} } + 2 {x}^{ - 3} + 12 = \\ = \frac{1}{3 \sqrt[3]{ {x}^{2} } } + \frac{15}{4 \sqrt[4]{x} } + \frac{2}{ {x}^{3} } + 12$

$y' = - 45 \sin(9x) - 2x \sin(2x) - 2 {x}^{2} \cos(2x) + \frac{1}{4}$

$y '= \frac{1}{4} {(4x - 23)}^{ - \frac{3}{4} } \times 4 { ln }^{7} (3 {x}^{2} - 7x) + 7 { ln }^{6} (3 {x}^{2} - 7x) \times (6x - 7) \times \sqrt{4x - 23} = \\ = \frac{ { ln }^{2}( 3 {x}^{2} - 7x)}{ \sqrt[4]{ {(4x - 23)}^{3} } } + 7(6x - 7) \sqrt{4x - 23} \times { ln}^{6} (3 {x}^{2} - 7x)$

$y' = \frac{2(tg(2x) + 10x) \times ( \frac{2}{ { \cos }^{2}(2x)} + 10) \times (3x + 2) - 3 {(4tg(2x) + 10x)}^{2} }{ {(3x + 2)}^{2} } = \\ = \frac{(tg(2x) + 10x)((6x + 4)( \frac{2}{ { \cos}^{2}(2x) } + 10) - 3(4tg(2x) + 10x)}{ {(3x + 2)}^{2} }$

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

$y' = 3 {e}^{tg(5t)} \times \frac{5}{ { \cos}^{2} (5t)} - 8 {t}^{3} {e}^{ \sin(6t) } - (2 {t}^{4} + 3) {e}^{ \sin(6t) } \times 6 \cos(6t) + 13 {(4t - 12)}^{12} \times 4 = \\ = \frac{15 {e}^{tg(5t)} }{ { \cos}^{2}(5t) } - {e}^{ \sin(6t) } (8 {t}^{3} + 6(2 {t}^{4} + 3) \cos(6t) ) + 52 {(4t - 12)}^{12}$

$y' = 3 {a}^{2} { ln}^{2} (3x - 7) \times 3 + 3 ln( \sin(5x) ) + 3x \times \frac{1}{ \sin(5x) } \times 5 \cos(5x) + 48x = \\ = 9 {a}^{2} { ln}^{2} (3x - 7) + 3 ln( \sin(5x) ) + 15xctg(5x) + 48x$

$q'(p) = 30 \cos(6p) \times {3}^{p} + ln(3) \times {3}^{p} \times 5 \sin(6p) + 5 { \cos}^{4} (13p - 5) \times ( - \sin(13p - 5)) \times 13 = \\ = {3}^{p} (30 \cos(6p) + 5 ln(3) \sin(6p) ) - 65 \sin(13p - 5) { \cos}^{4} (13p - 5)$

$y' = 3 {tg}^{2} ( \sin(2x) \cos(2x) ) \times \frac{1}{ { \cos }^{2} ( \sin(2x) \cos(2x) )} \times (2 \cos(2x) \cos(2x) - 2 \sin(2x) \sin(2x)) + 4 { ln}^{3} ( {e}^{3x} + 4 {x}^{2} ) \times \frac{1}{ {e}^{3x} + 4 {x}^{2} } \times (3 {e}^{3x} + 8x) = \\ = \frac{3 {tg}^{2} ( \sin(2x) \cos(2x)) \times 2 \cos(4x) }{ { \cos }^{2}( \sin(2x) \cos(2x)) } + \frac{4(3 {e}^{3x} + 8x) { ln}^{3}(3 {e}^{3x} + 8x) }{ {e}^{3x} + 4 {x}^{2} } = \\ = \frac{6 {tg}^{2} ( \sin(2x) \cos(2x) ) \times \cos(4x) }{ { \cos }^{2} ( \sin(2x) \cos(2x)) } + \frac{4(3 {e}^{3x} + 8x) { ln }^{3}( {e}^{3x} + 4 {x}^{2} ) }{ {e}^{3x} + 4 {x}^{2} }$

10.

$y '= 2 \sin(2y) + 2(2x + 3) \cos(2y) - 8 \cos(2y) + \frac{4}{3 {x}^{2} } \times 6x + 30 {x}^{2} = \\ = 2 \sin(2y) + (4x + 6) \cos(2y) - 8 \cos(2y) + \frac{8}{x} + 30 {x}^{2}$