Question

1) Найдите производные первого и второго порядка 2) найдите производные первого и второго порядка . Напишите уравнение касательной в точке х0=2
1. У= х^3+6х^2+9х
2. У= х^2/х-2

Answer 1

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

$y'' = 20 {x}^{3} - 2x + 2 - {x}^{ - 2} = \\ = 20 {x}^{3} - 2x + 2 - \frac{1}{ {x}^{2} }$

в х0 = 2

$y'(2) = 5 \times {2}^{4} - {2}^{2} + 4 + 5 + \frac{1}{2} = \\ = 80 - 4 + 4 + 5.5 = 85.5$

1.

$y = {x}^{3} + 6 {x}^{2} + 9x \\ y' = 3 {x}^{2} + 12x + 9 \\ y ''= 6x + 12$

2.

$y= \frac{ {x}^{2} }{x - 2} \\ y' = \frac{2x(x - 2) - {x}^{2} }{ {(x - 2}^{2} } = \frac{2 {x}^{2} - 4x - {x}^{2} }{ {(x - 2)}^{2} } = \\ = \frac{ {x}^{2} - 4x }{ {(x - 2)}^{2} } \\ y'' = \frac{(2x - 4) {(x - 2)}^{2} - 2(x - 2)( {x}^{2} - 4x)}{ {(x - 2)}^{4} } = \\ = \frac{(x - 2)((2x - 4)(x - 2) - 2( {x}^{2} - 4x) }{ {(x - 2)}^{4} } = \\ = \frac{2 {x}^{2} - 4x - 4x + 8 - 2 {x}^{2} + 8x }{ {(x - 2)}^{3} } = \\ = \frac{8}{ {(x - 2)}^{3} }$