Question

Найдите производную.
первый вариант, заранее

Answer 1

2 вариант:

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

$2)f'(x) = 8 \times \frac{1}{4} {x}^{ - \frac{3}{4} } = \frac{2}{ \sqrt[4]{ {x}^{3} } }$

$3)f'(x) = 5 \times ( - 7) {x}^{ - 8} = - \frac{ 35}{ {x}^{8} }$

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

$5)f'(x) = 7 {(4 - 3x)}^{6} \times ( - 3) = - 21 {(4 - 3x)}^{6}$

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

$7)f'(x) = \frac{1}{2 \sqrt{2x + 1} } \times 2 \times {(3x - 1)}^{4} + 4 {(3x - 1)}^{3} \times 3 \times \sqrt{2x + 1} = {(3x - 1)}^{3} ( \frac{3x - 1}{ \sqrt{2x + 1} } + 12 \sqrt{2x + 1} )$

$8)f'(x) = 16x {(3 - x)}^{4} + 4 {(3 - x)}^{3} \times ( - 1) \times 8 {x}^{2} = {(3 - x)}^{3} (16x(3 - x) - 32 {x}^{2} ) = {(3 - x)}^{3} (48x - 16 {x}^{2} - 32 {x}^{2} ) = {(3 - x)}^{3} (48x - 48 {x}^{2} )$

$9)f'(x) = \frac{4 {x}^{3}( {x}^{3} + 1) - 3 {x}^{2}( {x}^{4} - 1) }{ {( {x}^{3} + 1)}^{2} } = \frac{4 {x}^{6} + 4 {x}^{3} - 3 {x}^{6} + 3 {x}^{2} }{ {( {x}^{3} + 1)}^{2} } = \frac{ {x}^{6} + 4 {x}^{3} + 3 {x}^{2} }{ {( {x}^{3} + 1)}^{2} }$