Найдите производные функций!

Алгебра

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Vitalevich1799

22.08.2020

$1)y' = 6 {x}^{5}$

$2)y' = 4 - {x}^{ - 2} = 4 - \frac{1}{ {x}^{2} }$

$3)y' = \frac{1}{2} {x}^{ - \frac{1}{2} } - 4x = \frac{1}{2 \sqrt{x} } - 4x$

$4)y' = {x}^{3} - \frac{8}{2 \sqrt{x} } + 4 {x}^{ - 2} = \\ = {x}^{3} - \frac{4}{ \sqrt{x} } + \frac{4}{ {x}^{2} }$

$5)y' = 2(4 {x}^{2} - 3x + 7) + (8x - 3)(2x + 3) = \\ = 8 {x}^{2} - 6x + 14 + 16 {x}^{2} + 24x - 6x - 9 = \\ = 2 {x}^{2} + 12x + 5$

$5)y' = \frac{(2x - 2)(x + 4 {x}^{3}) - (1 + 12 {x}^{2} )( {x}^{2} - 2x) }{ {(x + 4 {x}^{3}) }^{2} } = \\ = \frac{2 {x}^{2} + 8 {x}^{4} - 2x - 8 {x}^{3} - {x}^{2} - 2x - 12 {x}^{4} + 24 {x}^{3} }{ {(x + 4 {x}^{3} )}^{2} } = \\ = \frac{ - 4 {x}^{4} + 16 {x}^{3} + {x}^{2} - 4x }{ {x}^{2} {(1 + 4 {x}^{2} )}^{2} } = \frac{ - 4 {x}^{3} + 16 {x}^{2} + x - 4 }{x {(1 + 4 {x}^{2}) }^{2} }$