Необходимо решить карточку за сегодняшний день, желательно быстрее )

Question

Алгебра

Геннадий

5 год назад

Необходимо решить карточку за сегодняшний день, желательно быстрее )

ОТВЕТЫ

Николай

Nov 24, 2020

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

$2)y' =-\frac{15}{3}{x}^{14}=- 5 {x}^{14}$

$3)y' =- 6 {x}^{2}+ 12 \times\frac{1}{2}{x}^{ -\frac{1}{2} }=- 6 {x}^{2}+\frac{6}{ \sqrt{x} }$

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

$5)y' = 1 \times{x}^{3}+ 3 {x}^{2} (x - 6) ={x}^{3}+ 3 {x}^{3}- 18 {x}^{2}= 4 {x}^{3}- 18 {x}^{2}$

$6)y' = ( {x}^{ \frac{5}{2}}+ 1)' =\frac{5}{2}{x}^{ \frac{3}{2} }= 2.5x \sqrt{x}$

$7)y' =\frac{1}{2}{(6x + 1)}^{ -\frac{1}{2} } \times 6 \times( {x}^{4}- 5) + 4 {x}^{3}\sqrt{6x + 1}=\frac{3({x}^{4} - 5) }{ \sqrt{6x + 1} }+ 4 {x}^{3}\sqrt{6x + 1}$

$8)y' ={( \frac{x}{3}+ 1)}^{3}+ 3x {( \frac{x}{3} + 1) }^{2}\times\frac{1}{3}= {( \frac{x}{3}+ 1)}^{3}+ x {( \frac{x}{3} + 1) }^{2}$

$9)y' =\frac{2(3 - 2x) - ( - 2)(2x + 3)}{ {(3 - 2x)}^{2} }=\frac{6 - 4x + 4x + 6}{ {(3 - 2x)}^{2} }=\frac{12}{ {(3 - 2x)}^{2} }$

$10)y' =\frac{3 {x}^{2}(2x - 3) - 2 {x}^{3}}{ {(2x - 3)}^{2} }=\frac{6 {x}^{3} - 9 {x}^{2}- 2 {x}^{3} }{ {(2x - 3)}^{2} }=\frac{4 {x}^{3} - 9 {x}^{2}}{ {(2x - 3)}^{2} }$

$11)y' =\frac{(4 {x}^{3} + 2 {x}^{2})(x + 1) -{x}^{4}-{x}^{2} - 1}{ {(x + 1)}^{2} }=\frac{4 {x}^{4}+ 4 {x}^{3} + 2 {x}^{3}+ 2 {x}^{2}-{x}^{4} -{x}^{2} - 1}{ {(x + 1)}^{2} }=\frac{3 {x}^{4}+ 6 {x}^{3}+{x}^{2} - 1 }{ {(x + 1)}^{2} }$

$12)y' =\frac{2 {x}^{5} (3x - 2) - 3( \frac{ {x}^{6} }{3}+ 2)}{ {(3x - 2)}^{2} }=\frac{6 {x}^{6}- 4 {x}^{5} -{x}^{6}- 6 }{ {(3x - 2)}^{2} }=\frac{5 {x}^{6}- 4 {x}^{5}- 6}{ {(3x - 2)}^{2} }$

$13)y' =\frac{15 {x}^{2} {(x - 4) - 2(x - 4) }^{2}5 {x}^{3}}{ {(x - 4)}^{4} }=\frac{(x - 4)(15 {x}^{2} - 10 {x}^{4} - 40 {x}^{3} )}{ {(x - 4)}^{4} }=\frac{15 {x}^{2} - 10 {x}^{4} - 40 {x}^{3}}{ {(x - 4)}^{3} }$

$14)y' =\frac{2x( {x}^{3}- x) - (3 {x}^{2}- 1) {x}^{2} }{ {( {x}^{3}- x)}^{2} }=\frac{2 {x}^{4}- 2 {x}^{2}- 3 {x}^{4} +{x}^{2}}{ {( {x}^{3}- x)}^{2} }=\frac{ -{x}^{4} -{x}^{2}}{ {( {x}^{3} - x) }^{2} }=\frac{ {x}^{2}( -{x}^{2}- 1) }{ {x}^{2} {( {x}^{2} - 1) }^{2}}=\frac{ -{x}^{2} - 1 }{ {x}^{2} - 1 }$

$15)y' = ((3x - 6 -{x}^{2}+ 2x)( {x}^{2}+ 2x + 3x + 6))' = (( -{x}^{2}+ 5x - 6)( {x}^{2}+ 5x + 6))' = ( - 2x + 5)( {x}^{2}+ 5x + 6) + (2x + 5)( -{x}^{2}+ 5x - 6) =- 2 {x}^{3}- 10 {x}^{2}- 12x + 5 {x}^{2}+ 25x + 30 - 2 {x}^{3}+ 10 {x}^{2}- 12x - 5 {x}^{2}+ 25x - 30 =- 4 {x}^{3}+ 10 {x}^{2}+ 26x$

$16)f'(x) =\frac{2}{ \sqrt{x} }-\frac{1}{10 {x}^{2} }$

$f'( \frac{1}{9}) =\frac{2}{ \sqrt{ \frac{1}{9} } }-\frac{1}{10 \times\frac{1}{81} }= 2 \times 3 -\frac{81}{10}= 6 -\frac{81}{10}=\frac{60 - 81}{10}=-\frac{21}{10}=- 2.1$

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