Question

Допоможіть, будь ласка, ДАЮ 100 БАЛІВ. НЕПРАВИЛЬНО БУДУ БАНИТЬ

Answer 1

Ответ:

1.

$f'(x) = 2 \times 3 {x}^{2} - \frac{1}{3} \times 3 {x}^{2} + 3 \times 2x - 0 = \\ = 6 {x}^{2} - {x}^{2} + 6x$

2.

$f'(x) = (3x - 5) '\times \sqrt{x} + ( {x}^{ \frac{1}{2} } )'(3x - 5) = \\ = 3 \sqrt{x} + \frac{1}{2} {x}^{ - \frac{1}{2} } (3x - 5) = \\ = 3 \sqrt{x} + \frac{1}{2 \sqrt{x} } (3x - 5) = \\ = 3 \sqrt{x} + 1.5 \sqrt{x} - \frac{5}{2 \sqrt{x} } = \\ = 4.5 \sqrt{x} - \frac{5}{2 \sqrt{x} }$

3.

$f'(x) = \frac{( {x}^{2} + 9x)'(x - 4) - (x - 4)'(x {}^{2} + 9x)}{ {(x - 4)}^{2} } = \\ = \frac{(2x + 9)(x - 4) - 1 \times ( {x}^{2} + 9x) }{ {(x - 4)}^{2} } = \\ = \frac{2 {x}^{2} - 8x + 9x - 36 - {x}^{2} - 9x }{ {(x - 4)}^{2} } = \\ = \frac{ {x}^{2} - 8x - 36 }{ {(x - 4)}^{2} }$

4.

$f'(x) = (2 {x}^{ - 3} - 3 {x}^{ - 6} ) '= 2 \times ( - 3) {x}^{ - 4} - 3 \times ( - 6) {x}^{ - 7} = \\ = - \frac{6}{ {x}^{4} } + \frac{18}{ {x}^{7} }$