Найдите производные следующих функций при х=1 1) (x^6-7)
2) (4x+1/x)
3) (корень из x-2x^2)
4(1/4x^4-8√x-4/x)
5) ((2x+3) (4x^2-3x+7))
6) (x^2-2x/x+4x^3)

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Ответ:

frezoloto

20.02.2021 09:02

$( {x}^{6} - 7)' = 6 {x}^{5} \\ \\x =1 \\ 6*1=6$

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

$( \sqrt{x } -2 {x}^{2} )' = \frac{1}{2} {x}^{-\frac{1}{2}} -4x=\\= \frac{1}{2\sqrt{x}}-4x\\\\x=1\\\frac{1}{2}-4=-3,5$

$( \frac{1}{4 {x}^{4} } - \frac{8}{ \sqrt{x} } - \frac{4}{x} )' = ( \frac{1}{4} {x}^{ - 4} - 8 {x}^{ - \frac{1}{2} } - 4 {x}^{ - 1} ) '= \\ = \frac{1}{4} \times ( - 4) {x}^{ - 5} - 8 \times ( - \frac{1}{2} ) {x}^{ - \frac{3}{2} } + 4 {x}^{ - 2} = \\ = - \frac{1}{ {x}^{5} } + \frac{4}{x \sqrt{x} } + \frac{4}{ {x}^{2} }\\ \\x=1\\-1+4+4=7$

$(2x + 3)'(4 {x}^{2} - 9x + 7) + (4 {x}^{2} - 9x + 7)'(2x + 3) = \\ = 2(4 {x}^{2} - 9x + 7) + (8x - 9)(2x + 3) = \\ = 8 {x}^{2} - 18x + 14 + 16 {x}^{2} + 24x - 18x - 24 = \\ = 24 {x}^{2} - 12x - 10\\\\x=1\\24-12-10=2$

$\frac{( {x}^{2} - 2x)'(x + 4 {x}^{3}) - (4 {x}^{3} + x)'( {x}^{2} - 2x) }{ {(4 {x}^{3} + x) }^{2} } = \\ = \frac{(2x - 2)(4 {x}^{3} + x) - (12 {x}^{2} + 1)( {x}^{2} - 2x)}{ {(4 {x}^{3} + x) }^{2} } = \\ = \frac{8 {x}^{4} + 2 {x}^{2} - 8 {x}^{3} - 2x - 12 {x}^{4} + 24 {x}^{3} - {x}^{2} + 2x}{ {(4 {x}^{3} + x) }^{2} } = \\ = \frac{ - 3 {x}^{4} + 16 {x}^{3} + {x}^{2} }{ {(4 {x}^{3} + x) }^{2} } = \frac{ {x}^{2}( - 3 {x}^{2} + 16x + 1) }{ {x}^{2} \times {(4 {x}^{2} + 1)}^{2} } = \\ = \frac{16x - 3 {x}^{2} + 1}{ {(4 {x}^{2} + 1)}^{2} }\\\\x=1\\ \frac{16 - 3 + 1}{ {(4 + 1)}^{2} }=\frac{14}{25}$