Question

Помогите решить.. Очень срочно..

Answer 1

Ответ:

$1)y' = 12 {x}^{5} - 32 {x}^{3} + 12 {x}^{2} + 4x$

$2)y' = \frac{5}{x} + 4 \sin(x) - 1$

$3)y' = 9 \sin(x) + (9x + 6) \cos(x)$

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

$5)y' = \cos(1 - 6x) \times ( - 6) = - 6 \cos(1 - 6x)$

$6)y' = 2x {e}^{ {x}^{2} + 2}$

$7)y' = \frac{5x}{x - 7} + 5 ln(x - 7) + 36 \sin(9x)$

$8)y' = 3 {x}^{2} log_{2}( {x}^{3} + x ) + {x}^{3} \times \frac{1}{ ln(2) \times ( {x}^{3} + x) } \times (3 {x}^{2} + 1) = 3 {x}^{2} log_{2}( {x}^{3} + x) + \frac{ {x}^{3}(3 {x}^{2} + 1) }{ ln(2) \times x( {x}^{2} + 1)} = 3 {x}^{2} log_{2}( {x}^{3} + x) + \frac{ {x}^{2}(3 {x}^{2} + 1) }{( {x}^{2} + 1) ln(2) }$

$9)y' = \frac{5 \cos(x - 4) + (7 + 5x) \sin(x - 4) }{ { \cos(x - 4) }^{2} }$

$10)y' = 5 {(2x + 1)}^{4} \times 2 \times {(x + 4)}^{9} + 9 {(x + 4)}^{8} {(2x + 1)}^{5} = {(2x + 1)}^{4} {(x + 4)}^{8} (10(x + 4) + 9(2x + 1)) = {(2x + 1)}^{4} {(x + 4)}^{8}(10x + 40 + 18x + 9) = {(2x + 1)}^{4} {(x + 4)}^{8}(28x + 49)$

$11)y' = \frac{ - ln(2) \times {2}^{ - x}(5x + ln(3x + 8)) - (5 + \frac{3}{3x + 8} ) {2}^{ - x} }{ {(5x + ln(3x + 8)) }^{2} } = \frac{ {2}^{ - x} ( - ln(2)(5x + ln(3x + 8) - (5 + \frac{3}{3x + 8} )) }{ {(5x + ln(3x + 8)) }^{2} }$