Question

Вариант 2 № 1. Найдите производную функции:
а) у = 8х+6х2 – 4х + 0,5 , вычислите значение производной в точке хо=3;
б) у = -2 cosx - х.
2x² – 3x
в) у =
х-х
г) у = (2x+3) - (5х2 - 2x2 - 1).
№2. Найдите производную функции:
а) у = 6х + 4х2 - 2,5х + 3 , вычислите значение производной в точке хо=4;
б) у = 5 sinx - x2,
3х2 - 10
в) y=
- x
г) у = (5х + 3) - (х – 6x? - 5).
2х4​

Answer 1

1.

а)

$y '= 8 \times 3 {x}^{2} + 6 \times 2x - 4 + 0 = \\ = 24 {x}^{2} + 12x - 4$

б)

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

в)

$y' = \frac{(2 {x}^{2} - 3x)' \times ( {x}^{3} - x) - ( {x}^{3} - x) '\times (2 {x}^{2} - 3x) }{ {( {x}^{3} - x)}^{2} } = \\ = \frac{(4x - 3)( {x}^{3} - x) - (3 {x}^{2} - 1)(2 {x}^{2} - 3x)}{ {( {x}^{3} - x)}^{2} } = \\ = \frac{4 {x}^{4} - 4 {x}^{2} - 3 {x}^{3} + 3x - 6 {x}^{4} + 9 {x}^{3} + 2 {x}^{2} - 3x }{ {( {x}^{3} - x) }^{2} } = \\ = \frac{ - 2 {x}^{4} + 6 {x}^{3} - 2 {x}^{2} }{ {(x( {x}^{2} - 1)) }^{2} } = \frac{ {x}^{2}( - 2 {x}^{2} + 6x - 2) }{ {x}^{2} {( {x}^{2} - 1) }^{2} } = \\ = - \frac{2 {x}^{2} - 6x + 2 }{ {( {x}^{2} - 1) }^{2} }$

г)

$y' = (2x + 3)'(5 {x}^{3} - 2 {x}^{2 } - 1) + (5 {x}^{3} - 2 {x}^{2} - 1)'(2x + 3) = \\ = 2(5 {x}^{3} - 2 {x}^{2} - 1) + (15 {x}^{2} - 4x)(2x + 3) = \\ = 10 {x}^{3} - 4 {x}^{2} - 2 + 30 {x}^{3} + 45 {x}^{2} - 8 {x}^{2} - 12x = \\ = 40 {x}^{3} + 33 {x}^{2} - 12x - 2$

2.

а)

$y '= 18 {x}^{2} + 8x - 2.5$

б)

$y' = (5 \sin(x)) ' \times {x}^{2} + ( {x}^{2} ) '\times 5 \sin(x) = \\ = 5 \cos(x) \times {x}^{2} + 2x \times 5 \sin(x) = \\ = 5 {x}^{2} \cos(x) + 10x \sin(x)$

в)

$y' = \frac{(3 {x}^{2} - 10)'(2 {x}^{4} - x) - (2 {x}^{4} - x)'(3 {x}^{2} - 10) }{ {(2 {x}^{4} - x) }^{2} } = \\ = \frac{6x(2 {x}^{4} - x) - (8 {x}^{3} - 1)(3 {x}^{2} - 10) }{ {(2 {x}^{4} - x) }^{2} } = \\ = \frac{12 {x}^{5} - 6 {x}^{2} - 24 {x}^{5} + 80 {x}^{3} + 3 {x}^{2} - 10 }{ {(x(2 {x}^{3} - 1)) }^{2} } = \\ = \frac{ - 12 {x}^{5} + 80 {x}^{3} - 3 {x}^{2} - 10}{ {x}^{2} {(2 {x}^{3} - 1) }^{2} } = - \frac{12 {x}^{5} - 80 {x}^{3} + 3 {x}^{2} + 10 }{ {x}^{2} {(2 {x}^{3} - 1)}^{2} }$

г)

$y' = (5x + 3)'( {x}^{3} - 6 {x}^{2} - 5) + ( {x}^{3} - 6 {x}^{2} - 5)'(5x + 3) = \\ = 5( {x}^{3} - 6 {x}^{2} - 5) + (3 {x}^{2} - 12x)(5x + 3) = \\ = 5 {x}^{3} - 30 {x}^{2} - 25 + 15 {x}^{3} + 9 {x}^{2} - 60 {x}^{2} - 36x = \\ = 20 {x}^{3} - 81x {}^{2} - 36x - 25$