Помогите пожалуйста! Срочно!. Created by demakefirchic33. matematika-ru

Ответ:87912345688012348811234

Помогите пожалуйста! Срочно!...

demakefirchic33 @demakefirchic33

September 2021 1 12 Report

Помогите пожалуйста! Срочно!

Answers & Comments

Miroslava227

Ответ:

879

$y' = 2 \cos(3x) \times ( \cos(3x)) ' \times (3x) '= \\ = 2 \cos(3x) \times ( - \sin(3x) ) \times 3 = - 3 \sin(6x)$

$y' = ( \sin(x) \cos(x)) ' + 1 = (\frac{1}{2} \times 2 \sin(x) \cos(x) ) '+ 1 = \\ = ( \frac{1}{2} \sin(2x) ) '+ 1 = \frac{1}{2} \cos(2x) \times 2 + 1 = \cos(2x) + 1$

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

$y '= 2 \sin( \frac{x}{2} ) \times \cos( \frac{x}{2} ) \times \frac{1}{2} = \frac{1}{2} \sin(x) \\$

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

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

880

$y '= \frac{(1 - \cos(2x)) '\times (1 + \cos(2x)) - (1 + \cos(2x)) '\times (1 - \cos(2x)) }{ (1 + \cos(2x)) {}^{2} } = \\ = \frac{ \sin(2x) \times 2(1 + \cos(2x)) + 2\sin(2x) (1 - \cos(2x)) }{(1 + \cos(2x)) {}^{2} } = \\ = \frac{ 2\sin(2x)(1 + \cos(2x) + 1 - \cos(2x)) }{ {(1 + \cos(2x)) }^{2} } = \\ = \frac{2 \sin(2x) \times 2 }{(1 + \cos {}^{2} (x) - \sin {}^{2} (x) ) {}^{2} } = \frac{4 \sin(2x) }{4 \cos {}^{4} (x) } = \\ = \frac{8 \sin(x) \cos(x) }{ 4\cos {}^{4} (x) } = \frac{2 \sin(x) }{ \cos {}^{3} (x) }$

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

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

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

881

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

$(( log_{2}(x) ) {}^{3} ) '= 3( log_{2}(x)) {}^{2} \times \frac{1}{ ln(2) x} = \\ = \frac{3 log_{2} {}^{2} (x) }{ ln(2) \times x }$

$(\sin( log_{3}(x)) ) ' = \cos( log_{3}(x) ) \times \frac{1}{ ln(3) \times x} \\$

$( \cos( {3}^{x} ) )' = - \sin( {3}^{x} ) \times ln(3) \times {3}^{x} \\$

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