Question

срочнодам 35 баллов

Вычислить производную следующих функций:

1. а) 1/х+5х-2; б) х2+√х − 5х+12; в) 6х2−11х+7/х;


2. а) 2√х • (4−3х); б) х2 • (5х2+3х – 7); в) (9х−12) • (6−8х);

3.а) (3х-2)/(8+5х); б) х^2/(2х-1); в) (3-4х)/х^2 ; г)(1-7х)/(1-9х);

Answer 1

Ответ:

1.

а)

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

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

б)

$y = {x}^{2} + \sqrt{x} - 5x + 12$

$y' = 2x + \frac{1}{2} {x}^{ - \frac{1}{2} } - 5 = \\ = 2x + \frac{1}{2 \sqrt{x} } - 5$

в)

$y = 6 {x}^{2} - 11x + \frac{7}{x} = 6 {x}^{2} - 11x + 7 {x}^{ - 1}$

$y' = 12x + 11 - 7 {x}^{ - 2} = \\ = 12x + 11 - \frac{7}{ {x}^{2} }$

2.

а)

$y = 2 \sqrt{x} (4 - 3x) = \\ = 8 \sqrt{x} - 6x \sqrt{x} = \\ = 8 {x}^{ \frac{1}{2} } - 6 {x}^{ \frac{3}{2} }$

$y' = 8 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 6 \times \frac{3}{2} {x}^{ \frac{1}{2} } = \\ = \frac{4}{ \sqrt{x} } - 9 \sqrt{x}$

б)

$y = {x}^{2} (5 {x}^{2} + 3x - 7) = \\ = 5 {x}^{4} + 3 {x}^{3} - 7 {x}^{2}$

$y' = 20 {x}^{3} + 9 {x}^{2} - 14x$

в)

$y = (9x - 12)(6 - 8x)$

$y' = 9(6 - 8x) - 8(9x - 12) = \\ = 54 - 72x - 72x + 96 = - 144x + 150$

3.

а)

$y = \frac{3x - 2}{8 + 5x}$

$y' = \frac{3(8 + 5x) - 5(3x - 2)}{ {(5x + 8)}^{2} } = \\ = \frac{24 + 15x - 15x + 10}{ {(5x + 8)}^{2} } = \\ = \frac{34}{ {(5x + 8)}^{2} }$

б)

$y = \frac{ {x}^{2} }{2x - 1}$

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

в)

$y = \frac{3 - 4x}{ {x}^{2} }$

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

г)

$y = \frac{1 - 7x}{1 - 9x}$

$y' = \frac{ - 7(1 - 9x) + 9(1 - 7x)}{ {(1 - 9x)}^{2} } = \\ = \frac{ - 7 + 63x + 9 - 63x}{ {(1 - 9x)}^{2} } = \\ = \frac{2}{ {(1 - 9x)}^{2} }$