Question

Знайдіть похідну функції:
СРОЧНО!!!!!!!​

Answer 1

1

$f'(x) = \frac{1}{3} \times 6 {x}^{5} - 8 \times \frac{1}{2} {x}^{ - \frac{1}{2} } + 2 + 0 = \\ = 2 {x}^{5} - \frac{4}{ \sqrt{x} } + 2$

2

$f'(x) = (2 {x}^{ - 2} - 3 {x}^{ - 3} ) = \\ = 2 \times ( - 2) {x}^{ - 3} - 3 \times ( - 3) {x}^{ - 4} = - \frac{4}{ {x}^{3} } + \frac{9}{ {x}^{4} }$

3

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

4

$f'(x) = ( {x}^{3} - 2)'( {x}^{2} + 1) + ( {x}^{2} + 1)'( {x}^{3} - 2) = \\ = 3 {x}^{2} ( {x}^{2} + 1) + 2x( {x}^{3} - 2) = \\ = 4 {x}^{4} + 3 {x}^{2} + 2 {x}^{4} - 4x = 6 {x}^{4} + 3 {x}^{2} - 4x$

5

$f'(x) = (3x)'tgx + (tgx) '\times 3x = \\ = 3tgx + \frac{3x}{ \cos {}^{2} (x) }$

6

$f'(x) = \frac{( {x}^{2} + 5x)'(x - 3) - (x - 3)'( {x}^{2} + 5x) }{ {(x - 3)}^{2} } = \\ = \frac{(2x + 5)(x - 3) - ( {x}^{2} + 5x) }{ {(x - 3)}^{2} } = \\ = \frac{2 {x}^{2} - 6x + 5x - 15 - {x}^{2} - 5x }{ {(x - 3)}^{2} } = \\ = \frac{ {x}^{2} - 6x - 15 }{ {(x - 3)}^{2} }$