Question

Найти производные. Чем сможете помогите. Ставлю лучший

Answer 1

Ответ:

1)

$y' = 55 {x}^{10} - 18 {x}^{8} + 5 {x}^{4}$

2)

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

3)

$y' = ( {x}^{2} )' \sin(8x) + ( \sin(8x)) ' \times {x}^{2} = \\ = 2x \sin(8x) + \cos(8x) \times 8 \times {x}^{2} = \\ = 2x\sin(8x) + 8 {x}^{2} \cos(8x)$

4)

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