Question

Алгебра 10ый класс, оба варианта​

Answer 1

$1)y '= 3 {x}^{2}$

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

$3)y' = 25 {x}^{4} - \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 25 {x}^{4} - \frac{1}{2 \sqrt{x} }$

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

$5)y' = 3(7 {x}^{3} + 5x - 4) + (21 {x}^{2} + 5)(3x + 7) = \\ = 21 {x}^{3} + 15x - 12 + 63 {x}^{3} + 147 {x}^{2} + 15x + 35 = \\ = 84 {x}^{3} + 147 {x}^{2} + 30x + 23$

$6)y' = \frac{8x(5 - 2 {x}^{3}) + 6 {x}^{2} (4 {x}^{2} + 8)}{ {(5 - 2 {x}^{3}) }^{2} } = \\ = \frac{40x - 16 {x}^{4} + 24 {x}^{4} + 48 {x}^{2} }{ {(5 - 2 {x}^{3}) }^{2} } = \\ = \frac{8 {x}^{4} + 48 {x}^{2} + 40x }{ {(5 - 2 {x}^{3}) }^{2} }$

2 вариант

$1)y '= 5 {x}^{4}$

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

$3)y' = 16 {x}^{3} + \frac{1}{2 \sqrt{x} }$

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

$5)y' = 5(2 {x}^{4} - 7x + 1) + (8 {x}^{3} - 7)(5x - 4) = \\ = 10 {x}^{4} - 35x + 5 + 40 {x}^{4} - 32 {x}^{3} - 35x + 28 = \\ = 50 {x}^{4} - 32 {x}^{3} - 70x + 33$

$6)y' = \frac{3 {x}^{2} (3 - 4 {x}^{4} ) + 16 {x}^{3}( {x}^{3} - 7) }{ {(3 - 4 {x}^{4}) }^{2} } = \\ = \frac{9 {x}^{2} - 12 {x}^{6} + 16 {x}^{6} - 112 {x}^{3} }{ {(3 - 4 {x}^{4}) }^{2} } = \\ = \frac{4 {x}^{6} - 112 {x}^{3} + 9 {x}^{2} }{ {(3 - 4 {x}^{4} )}^{2} }$