только 16 числа больше не надо

Question

Геометрия

Ibse

3 year(s) ago

только 16 числа больше не надо

ответы: 1

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Ответ:

1.

$y' = 3 times 4 {x}^{3}= 12 {x}^{3}$

2.

$y' =frac{1}{2}times 6 {x}^{5} = 3 {x}^{5}\$

3.

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

4.

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

5.

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

6.

$y' = 20{x}^{3} - 14x - 1$

7.

$y' = 5 {x}^{4}+ 6 {x}^{2}+frac{ {x}^{ - 2} }{2}= 5 {x}^{4}+ 6 {x}^{2}+frac{1}{2 {x}^{2} }\$

8.

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

9.

$y = ((x +frac{1}{ sqrt{x} }-frac{1}{ {x}^{2} } )(3 {x}^{2}+ x - 8) )'=\= (3x {}^{3}+{x}^{2}- 8x + 3x sqrt{x}+sqrt{x}-frac{8}{ sqrt{x} }- 3 -frac{1}{x}+frac{8}{ {x}^{2} } ) '=\= (3 {x}^{3}+{x}^{2}- 8x + 3 {x}^{ frac{3}{2} }+{x}^{ frac{1}{2} }- 8 {x}^{ -frac{1}{2} }- 3 -{x}^{ - 1}+ 8 {x}^{ - 2} ) '=\= 9 {x}^{2}+ 2x - 8 +frac{9}{2}{x}^{ frac{1}{2} }+frac{1}{2}{x}^{ -frac{1}{2} }+ 4 {x}^{ -frac{3}{2} }- 0 +{x}^{ - 2}- 16 {x}^{ - 3}=\= 9 {x}^{2}+ 2x - 8 +frac{9}{2}sqrt{x}+frac{1}{2 sqrt{x} }+frac{4}{x sqrt{x} }+frac{1}{ {x}^{2} }-frac{16}{ {x}^{3} }$

10.

$y' = ((x +frac{1}{x}+frac{1}{ {x}^{2} } )( {x}^{2}+ x + 1) )'=\= ( {x}^{3}+{x}^{2}+ x + x + 1 +frac{1}{x}+ 1 +frac{1}{x}+frac{1}{ {x}^{2} } )' =\= ( {x}^{3}+{x}^{2}+ 2x + 2 +frac{2}{x}+frac{1}{ {x}^{2} } )' =\= 3 {x}^{2}+ 2x + 2 + 0 - 2 {x}^{ - 2}- 2 {x}^{ - 3}=\= 3 {x}^{2}+ 2x + 2 -frac{2}{ {x}^{2} }-frac{2}{ {x}^{3} }\$

11.

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

12.

$y' =frac{(x - 2)'(2 {x}^{2} - 1) - (2 {x}^{2}- 1)'(x - 2)}{ {(2 {x}^{2}- 1)}^{2} }=\=frac{1 times (2 {x}^{2}- 1) - 4x(x - 2)}{ {(x - 2)}^{2} }=frac{2 {x}^{2}- 1 - 4 {x}^{2}+ 8x}{ {(x - 2)}^{2} }=\=frac{ - 2x {}^{2} + 8x - 1 }{ {(x - 2)}^{2} }=-frac{2 {x}^{2}- 8x + 1}{ {(x - 2)}^{2} }$