Question

Розв'яжіть будь ласка!!!!! Дам 30 балів

Answer 1

Ответ:

а

$y' = 20 {x}^{19}$

б

$y' = 3 + 0 = 3$

в

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

г

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

д

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

е

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

ж

$y '= \frac{(3 - 4x)' \times {x}^{2} - ( {x }^{2} ) '\times (3 - 4x)}{( {x}^{2}) {}^{2} } = \\ = \frac{ - 4 \times {x}^{2} - 2x(3 - 4x) }{ {x}^{4} } = \\ = \frac{ - 4 {x}^{2} - 6x + 8 {x}^{2} }{ {x}^{4} } = \frac{4 {x}^{2} - 6x }{ {x}^{4} } = \\ = \frac{4x - 6}{ {x}^{3} }$

з

$y' = 3 \cos( \frac{x}{3} ) \times ( \frac{x}{3} ) '- \frac{1}{2} = \\ = 3 \cos( \frac{x}{3} ) \times \frac{1}{3} - 0.5 = \\ = \cos( \frac{x}{3} ) - 0.5$