Question

Срочно мне нужно до завтра, иначе я буду не аттестованым, мне нужен полный ответ??​

Answer 1

Ответ:

1)

$y = 2x {}^{5} + \frac{x {}^{3} }{3} + 1$

$y = \frac{d}{dx} (2x {}^{5} + \frac{x {}^{3} }{3} + 1)$

$y = \frac{d}{dx} (2x {}^{5} ) + \frac{d}{dx} ( \frac{x {}^{3} }{3} ) + \frac{d}{dx} (1)$

$2 \times \frac{d}{dx} (x {}^{5} )$

$2 \times 5x {}^{4}$

$y = 2 \times 5x {}^{4} + \frac{d}{dx} ( \frac{x {}^{3} }{3} ) + \frac{d}{dx} (1)$

$\frac{1}{3} \times \frac{d}{dx} (x {}^{3} )$

$\frac{1}{3} \times 3x {}^{2}$

$y = 2 \times 5x {}^{4} + \frac{1}{3} \times 3x {}^{2} + \frac{d}{dx} (1)$

$\frac{d}{dx} (1)$

$0$

$y = 2 \times 5x {}^{4} + \frac{1}{3} \times 3x {}^{2} + 0$

$2 \times 5x {}^{4} + \frac{1}{3} \times 3x {}^{2}$

Сокращаем на НОД 3:

$2 \times 5x {}^{4} + x {}^{2}$

$10x {}^{4} + {x}^{2}$

$y = 10x {}^{4} + x {}^{2}$

Ответ:в)

2)

X)

$y = 3tgx + \frac{7}{x}$

$\frac{dy}{dx} = \frac{d}{dx} (3tgx + \frac{7}{x} )$

$\frac{dy}{dx} = \frac{d}{dx} (3tgx) + \frac{d}{dx} ( \frac{7}{x} )$

$\frac{d}{dx} (3tgx)$

$3tg$

$\frac{dy}{dx} = 3tg + \frac{d}{dx} ( \frac{7}{x} )$

$\frac{d}{dx} ( \frac{7}{x} )$

$- 7 \times \frac{ \frac{d}{dx} (x)}{ {x}^{2} }$

$- 7 \times \frac{1}{ {x}^{2} }$

$\frac{dy}{dx} = 3tg - 7 \times \frac{1}{ {x}^{2} }$

$3gt - 7 \times \frac{1}{ {x}^{2} }$

$3gt - \frac{7}{ {x}^{2} }$

$\frac{dy}{dx} = 3gt - \frac{7}{ {x}^{2} }$

g)

$y = 3tgx + \frac{7}{x}$

$\frac{dy}{dx} = \frac{d}{dg} (3tgx + \frac{7}{x} )$

$\frac{dy}{dg} = \frac{d}{dg} (3txg + \frac{7}{x} )$

$\frac{dy}{dg} = \frac{d}{dg} (3txg) + \frac{d}{dx} ( \frac{7}{x} )$

$\frac{d}{dg} (3txg)$

$3tx$

$\frac{dy}{dg} = 3tx + \frac{d}{dg} ( \frac{7}{x} )$

$\frac{d}{dg} ( \frac{7}{x} )$

$0$

$\frac{dy}{dg} = 3tx + 0$

$\frac{dy}{dg} = 3tx$

t)

$y = 3tgx + \frac{7}{x}$

$\frac{dy}{dt} = \frac{d}{dt} (3tgx + \frac{7}{x} )$

$\frac{dy}{dt} = \frac{d}{dt} (3gxt + \frac{7}{x} )$

$\frac{dy}{dt} = \frac{d}{dt} (3gxt) + \frac{d}{dt} ( \frac{7}{x} )$

$\frac{d}{dt} (3gxt)$

$3gx$

$\frac{dy}{dt} = 3gx + \frac{d}{dt} ( \frac{7}{x} )$

$\frac{d}{dt} ( \frac{7}{x} )$

$0$

$\frac{dy}{dt} = 3gx + 0$

$\frac{dy}{dt} = 3gx$

Ответ:хз

3)

$y = \sqrt{x} \times \cos(x)$

$y = \frac{d}{dx} ( \sqrt{x} \cos(x) )$

$y = \frac{d}{dx} ( \sqrt{x} ) \times \cos(x) + \sqrt{x} \times \frac{d}{dx} ( \cos(x) )$

$\frac{d}{dx} ( \sqrt{x} )$

$\frac{1}{2 \sqrt{x} }$

$y = \frac{1}{2 \sqrt{x} } \times \cos(x) + \sqrt{x} \times \frac{d}{dx} ( \cos(x) )$

$\frac{d}{dx} ( \cos(x) )$

$- \sin(x)$

$y = \frac{1}{2 \sqrt{x} } \times \cos(x) + \sqrt{x} \times ( - \sin(x) )$

$\frac{ \cos(x) }{2 \sqrt{x} } + \sqrt{x} \times ( - \sin(x) )$

$\frac{ \cos(x) }{2 \sqrt{x} } - \sqrt{x} \sin(x)$

$\frac{ \cos(x) - 2 x \times \sin(x) }{2 \sqrt{x} }$