Question

решить 1 вариант. заранее огромное

Answer 1

$1)f'(x)= 8 {x}^{3} - 3 {x}^{2} + 3$

$2)f'(x) = 6 \times \frac{1}{3} {x}^{ - \frac{2}{3} } = \frac{2}{ \sqrt[3]{ {x}^{2} } }$

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

$4)f'(x) = \frac{2}{ \sqrt[3]{ {x}^{2} } } - \frac{2}{ {x}^{3} }$

$5)f'(x) = 8 {(2x + 3)}^{7} \times 2 = 16 {(2x + 3)}^{7}$

$6)f'(x) = \frac{1}{3} {(3x - 2)}^{ - \frac{2}{3} } \times 3 = \frac{1}{ \sqrt[3]{ {(3x - 2)}^{2} } }$

$7)f'(x) = 2(2x - 5) \times 2 \times \sqrt{x - 4} + \frac{1}{2 \sqrt{x - 4} } {(2x - 5)}^{2} = (2x - 5)(4 \sqrt{x - 4} + \frac{2x - 5}{2 \sqrt{x - 4} } )$

$8)f'(x) = 3 {(4 - x)}^{3} + 3x \times 3{(4 - x)}^{2} \times ( - 1) = {(4 - x)}^{2} (3(4 - x) - 9x) = {(4 - x)}^{2} (12 - 3x - 9x) = {(4 - x)}^{2} (12 - 12x)$

$9)f'(x) = \frac{3 {x}^{2}( {x}^{2} + 1) - 2x( {x}^{3} + 1) }{ {( {x}^{2} + 1) }^{2} } = \frac{3 {x}^{4} + 3 {x}^{2} - 2 {x}^{4} - 2x}{ {( {x}^{2} + 1) }^{2} } = \frac{ {x}^{4} + 3 {x}^{2} - 2x}{ {( {x}^{2} + 1)}^{2} }$