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Ответ:

Deelay

23.12.2020 15:26

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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