Question

Помогите решить а, б, в, г​

Answer 1

Ответ:

а

$y' = 4 \times 2x + \cos(2x) \times (2x)' - \frac{1}{ ln(5) \times x } = \\ = 8x + 2 \cos(2x) - \frac{1}{x ln(5) }$

б

$y '= ( ln(2x)) ' \times \cos(3x) + ( \cos(3x)) ' \times ln(2x) = \\ = \frac{1}{2x} \times (2x)' \times \cos(3x) - \sin(3x) \times (3x) '\times ln(2x) = \\ = \frac{1}{2x} \times 2 \cos(3x) - 3 \sin(3x) ln(2x) = \\ = \frac{ \cos(3x) }{x} - 3 \sin(3x) ln(2x)$

в

$y '= \frac{(3 - 5 {x}^{3})' \times ( {x}^{2} - 2) - ( {x}^{2} - 2)' \times (3 - 5 {x}^{3}) }{ {( {x}^{2} - 2)}^{2} } = \\ = \frac{ - 15 {x}^{2} ( {x}^{2} - 2) - 2x(3 - 5 {x}^{3} )}{( {x}^{2} - 2) {}^{2} } = \\ = \frac{ - 15 {x}^{4} + 30 {x}^{2} - 6x + 10 {x}^{4} }{( {x}^{2} - 2) {}^{2} } = \\ = \frac{ - 5 {x}^{4} + 30 {x}^{2} - 6x }{( {x}^{2} - 2) {}^{2} }$

г

$y '= \cos( {x}^{4} - 6 {x}^{3} + 11 ) \times ( {x}^{4} - 6 {x}^{3} + 11) '= \\ = \cos( {x}^{4} - 6 {x}^{3} + 11 ) \times (4 {x}^{3} - 18 {x}^{2} )$

Answer 2

а) у` = 8x + 4cos(2x) - 1/xln5

б) у` = 1/х + sin (3x)

в) у` = (-5х⁴+30х²-6х)/(х²-2)²

г) у` = cos (x⁴-6x³+11) × (4x³ - 18x²)