Алгебра: С 6 примера , с решением

С 6 примера
, с решением

С 6 примера , с решением

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Ответ:

kuzyaevkuzyaev2

03.04.2021 09:11

$1)(3 {x}^{5} + 7 {x}^{4} + 5 {x}^{2} + x - 3) '= \\ = 3 \times 5 {x}^{4} + 7 \times 4 {x}^{3} + 5 \times 2x + 1 = \\ = 15 {x}^{4} + 28 {x}^{3} + 10x + 1$

$2)(7 {x}^{4} + 28 {x}^{3} + 5 {x}^{2} + 125) '= \\ = 7 \times 4 {x}^{3} + 28 \times 3 {x}^{2} + 5 \times 2x + 0 = \\ = 28 {x}^{3} + 84 {x}^{2} + 10x$

$3)(3 {x}^{5} - 15 {x}^{4} + 60 {x}^{3} + 180 {x}^{2} + 360) '= \\ = 15 {x}^{4} - 60 {x}^{3} + 180 {x}^{2} + 360 x$

$4)(3 {x}^{7} - 15 {x}^{4} + 60 {x}^{2} + 120)' = \\ = 21 {x}^{6} - 60 {x}^{3} + 120x$

$5)( \frac{1}{ {x}^{4} } + 3 {x}^{5} + 153)' = ( {x}^{ - 4} + 3 {x}^{5} + 153) '= \\ = - 4 {x}^{ - 5} + 15 {x}^{4} + 0 = - \frac{4}{ {x}^{5} } + 15 {x}^{4}$

$6)( \frac{1}{ {x}^{5} } + \frac{1}{ {x}^{4} } + \frac{1}{ {x}^{3} } + \sqrt{x} ) '= \\ = ( {x}^{ - 5} + {x}^{ - 4} + {x}^{ - 3} + {x}^{ \frac{1}{2} } )' = \\ = - 5 {x}^{ - 6} - 4 {x}^{ - 5} + 3 {x}^{ - 4} + \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = - \frac{5}{ {x}^{6} } - \frac{4}{ {x}^{5} } + \frac{3}{ {x}^{4} } + \frac{1}{2 \sqrt{x} }$

$7)( {x}^{ - 6} + 3 {x}^{ - 5} + 6 {x}^{ - 4} + 9 {x}^{ - 3} )' = \\ = - 6 {x}^{ - 7} - 15 {x}^{ - 6} - 24 {x}^{ - 5} - 27 {x}^{ - 4} = \\ = - \frac{6}{ {x}^{7} } - \frac{15}{ {x}^{6} } - \frac{24}{ {x}^{5} } - \frac{27}{ {x}^{4} }$

$8)( \frac{3}{5} {x}^{ \frac{3}{5} } + 3 {x}^{3} + 5 {x}^{2} )' = \\ = \frac{3}{5} \times \frac{3}{5} {x}^{ - \frac{2}{5} } + 9 {x}^{2} + 10x = \\ = \frac{9}{25 \sqrt[5]{ {x}^{2} } } + 9 {x}^{2} + 10x$

$9)( {x}^{ \frac{4}{5 } } + {x}^{ \frac{3}{2} } + {x}^{ \frac{2}{3} } + {x}^{ \frac{1}{2} } + 3)' = \\ = \frac{4}{5} {x}^{ - \frac{1}{5} } + \frac{3}{2} {x}^{ \frac{1}{2} } + \frac{2}{3} {x}^{ - \frac{1}{3} } + \frac{1}{2} {x}^{ - \frac{1}{2} } + 0 = \\ = \frac{4}{5 \sqrt[5]{x} } + \frac{3}{2} \sqrt{x} + \frac{2}{3 \sqrt[3]{x} } + \frac{1}{2 \sqrt{x} }$

$10)(5 {x}^{ \frac{5}{4} } + 4 {x}^{ \frac{4}{3} } + 3 {x}^{ \frac{3}{2} } + 75)' = \\ = 5 \times \frac{5}{4} {x}^{ \frac{1}{4} } + 4 \times \frac{4}{3} {x}^{ \frac{1}{3} } + 3 \times \frac{3}{2} {x}^{ \frac{1}{2} } + 0 = \\ = \frac{25}{4} \sqrt[4]{x} + \frac{16}{3} \sqrt[3]{x} + \frac{9}{2} \sqrt{x}$