Предмет: Математика, автор: jekakrutov2016

решите примеры пожалуйста, я вас умоляю кто нибуть помогите!!

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

$y = 2 {x}^{2} - 4x \\ x_0 = 1$

$f(x) = y(x_0) + y'(x_0)(x - x_0)$

$y(0) = 2 - 4 = - 2$

$y' = 4x - 4$

$y'(1) = 4 - 4 = 0$

$f(x) = - 2 + 0 \times (x - 1) = - 2$

- уравнение касательной

$s(t) = {t}^{3} + 5 t {}^{2} - 3 \\ t = 2$

$v(t) = s'(t) = 3 {t}^{2} + 10 t - 3$

$v(2) = 3 \times 4 + 20 - 3 = 32 - 3 = 29$

$a(t) = v'(t) = 6t + 10 \\ a(2) = 12 + 10 = 22$

а)

$y '= - \sin(x) - 5 {x}^{ - 6} = - \sin(x) - \frac{5}{ {x}^{6} } \\$

б)

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

$y' = ({x}^{2} ) '\times \sin(x) + ( \sin(x)) ' \times {x}^{2} = \\ = 2x \sin(x) + {x}^{2} \cos(x) \\ \\ y'( \frac{\pi}{3} ) = \frac{2\pi}{3} \sin( \frac{\pi}{3} ) + \frac{ {\pi}^{2} }{9} \cos( \frac{\pi}{3} ) = \\ = \frac{2\pi}{3} \times \frac{ \sqrt{3} }{2} + \frac{ {\pi}^{2} }{9} \times \frac{1}{2} = \\ = \frac{ \sqrt{3} \pi}{3} + \frac{ {\pi}^{2} }{18}$

$f'(x) = \frac{( {x}^{2} + 15)'(x + 1) - (x + 1)'( {x}^{2} + 15)}{ {(x + 1)}^{2} } = \\ = \frac{2x(x + 1) - ( {x}^{2} + 15) }{ {(x + 1)}^{2} } = \frac{2 {x}^{2} + 2x - {x}^{2} - 15}{ {(x + 1)}^{2} } = \\ = \frac{ {x}^{2} + 2x - 15}{ {(x + 1)}^{2} } \\ \\ {x}^{2} + 2x - 15 =(x + 5)(x - 3) \\ D= 4 + 60 = 64 \\ x_1 = \frac{ - 2 + 8}{2} = 3 \\ x_2 = - 5 \\ \\ f(x) = \frac{( x + 5)(x - 3)}{ {(x + 1)}^{2} }$

$f'(x) = 0 \\ \\ \frac{(x + 5)(x - 3)}{ {(x + 1)}^{2} } = 0 \\ x_1 = - 5 \\ x_2 = 3 \\ x\ne - 1$

$f'(x) > 0 \\ \frac{(x + 5)(x - 3)}{ {(x + 1)}^{2} } > 0 \\ + \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: + \\ - -( - 5) - - ( - 1)- -3 - > \\ x\in( - \infty; - 5)U( 3 ;+ \infty )$

а)

$y = \frac{1}{(5x + 1) {}^{3} } = {(5x + 1)}^{ - 3} \\$

$y' = - 3 {(5x + 1)}^{ - 4} \times (5x + 1) '= \\ = \frac{ - 3}{ {(5x + 1)}^{4} } \times \times 5 = - \frac{15}{ {(5x + 1)}^{4} }$

б)

$y = \sqrt{ \sin(3x) } = ( \sin(3x)) {}^{ \frac{1}{3} }$

$y' = \frac{1}{3} ( \sin(3x)) {}^{ - \frac{2}{3} } \times ( \sin(3x)) '\times (3x) '= \\ = \frac{1}{3 \sqrt[3]{ \sin {}^{2} (3x) } } \times \cos(3x) \times 3 = \\ = \frac{ \cos(3x) }{ \sqrt[3]{ \sin {}^{2} (3x) } }$