Question

Найти производную следующих функций

Answer 1

Ответ:

$1)y- = 6x + 2$

$2)y- = (24 {x}^{2} + 3)(5x - 2) + 5(8 {x}^{3} + 3x - 1) = 120 {x}^{3} - 48 {x}^{2} + 15x - 6 + 40 {x}^{3} + 15x - 5 = 160 {x}^{3} - 48 {x}^{2} + 30x - 11$

$3)y' = \frac{(4x - 3)(4x + 1) - 4(2 {x}^{2} - 3x + 4) }{ {(4x + 1)}^{2} } = \frac{16 {x}^{2} + 4x - 12x - 3 - 8 {x}^{2} + 12x - 16}{ {(4x + 1)}^{2} } = \frac{8 {x}^{2} + 4x - 18 }{ {(4x + 1)}^{2} }$

$4)y' = 12 {x}^{2} + \frac{1}{2 \sqrt{x} } + 4 {x}^{ - 5} = 12 {x}^{2} + \frac{1}{2 \sqrt{x} } + \frac{4}{ {x}^{5} }$

$5)y' = 3 {(7 {x}^{4} + 4 {x}^{2} - 3) }^{2} \times (28 {x}^{3} + 8x)$

$6)y' = 2 {e}^{2x - 1} + \frac{5}{5x + 3}$

$7)y' = ln(5) \times {5}^{3x - 4} \times 3 + \frac{6}{ ln(3) \times (6x - 2)} = 3 ln(5) \times {5}^{3x - 4} + \frac{3}{ ln(3) \times (3x - 1) }$

$8)y' = \frac{1}{2 \sqrt{x + 1} } \times {e}^{4x - 2} + 4 {e}^{4x - 2} \sqrt{x + 1} = {e}^{4x - 2} ( \frac{1}{2 \sqrt{x + 1} } + 4 \sqrt{x + 1} )$

$9)y' = \cos(8x + 3) \times 8$

$10)y' = - \sin(2 - 4x) \times ( - 4) = 4 \sin(2 - 4x)$

$11)y' = - \frac{3}{ { \cos(3x) }^{2} } \times ctg(x) - \frac{1}{ { \sin(x) }^{2} } \times tg(3x)$

$12)y' = \frac{1}{ \sqrt{1 - 9 {x}^{2} } } \times 3$