Решите производные сложной функции!. Created by ЮжныйПарк. matematika-ru

Ответ:111213141516171819По формуле:20

Решите производные сложной функции!...

ЮжныйПарк @ЮжныйПарк

September 2021 1 9 Report

Решите производные сложной функции!

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Miroslava227

Ответ:

$y = ln( \cos(x) ) + 5 {e}^{ - 3x + 2} - 2$

$y '= \frac{1}{ \cos(x) } \times ( \cos(x)) ' + 5 {e}^{2 - 3x} \times (2 - 3x) '= \\ = \frac{1}{ \cos(x) } \times ( - \sin(x) ) + 5 {e}^{2 - 3x} \times ( - 3) = \\ = - tgx - 15 {e}^{2 - 3x}$

$y = (3x + 1) ln( {x}^{2} + 7) + 6$

$y '= (3x + 1)' ln( {x}^{2} + 7) + ( ln( {x}^{2} + 7))' (3x + 1) + 0 = \\ = 3 ln( {x}^{2} + 7 ) + \frac{1}{ {x}^{2} + 7} \times 2x(3x + 1) = \\ = 3 ln( {x}^{2} + 7) + \frac{6 {x}^{2} + 2x }{ {x}^{2} + 7}$

$y = (5 {x}^{3} - 4)arcsin(3x) + {3}^{x}$

$y '= 15 {x}^{2} \times arcsin(3x) + \frac{1}{ \sqrt{1 - 9 {x}^{2} } } \times 3 \times( 5 {x}^{3} - 4) + ln(3) \times {3}^{x} = \\ = 15 {x}^{2} arcsin(3x) + \frac{15 {x}^{3} - 12}{ \sqrt{1 - 9 {x}^{2} } } - ln(3) \times {3}^{x}$

$y = \frac{ {e}^{2x} }{ \sin( {x}^{2} - 1) } \\$

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

$y = \frac{ ln( \cos(x) ) }{ ln( 3 {x}^{4} + 1 ) } \\$

$y' = \frac{ \frac{1}{ \cos(x) } \times ( - \sin(x)) \times ln(3 {x}^{4} + 1) - \frac{1}{3 {x}^{4} + 1} \times 12 {x}^{3} ln( \cos(x) ) }{ {ln}^{2} (3 {x}^{4} + 1) } = \\ = \frac{ - tgx ln(3 {x}^{4} + 1) - \frac{12 {x}^{3} ln( \cos(x) ) }{3 {x}^{4} + 1} }{ {ln}^{2} (3 {x}^{4} + 1)}$

$y = {tg}^{3} ( \cos( \sqrt{x} ) )$

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

$y = {e}^{ctg2x} \sin(3x)$

$y' = {e}^{ctg2x} \times ( - \frac{1}{ \sin {}^{2} (2x) } ) \times 2 \times \sin(3x) + 3 \cos(3x) {e}^{ctg2x} = \\ = {e}^{ctg2x} (3 \cos(3x) - \frac{2 \sin(3x) }{ \sin {}^{2} (2x) } )$

$y = {6}^{2x - 1} \cos( \sqrt{4 - x} )$

$y '= ln(6) \times {6}^{2x - 1} \times 2 \cos( \sqrt{4 - x} ) + ( - \sin( \sqrt{4 - x} ) ) \times \frac{1}{2 \sqrt{4 - x} } \times ( - 1) \times {6}^{2x - 1} = \\ = {6}^{2x - 1} (2 ln(6) \times \cos( \sqrt{4 - x} ) + \frac{ \sin( \sqrt{4 - x} ) }{2 \sqrt{4 - x} } )$

$y = {( \sin(x)) }^{x}$

По формуле:

$y' = ( ln(y)) ' \times x$

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

$y '= {( \sin(x)) }^{x} ( ln( \sin(x) ) + xctg(x)) \\$

$y = {( {x}^{3} + 2) }^{5x}$

$( ln(y)) ' = ( ln( {x}^{3} + 2 ) {}^{5x} )' = \\ = (5x \times ln( {x}^{3} + 2)) '= \\ = 5 ln( {x}^{3} + 2) + 5x \times \frac{1}{ {x}^{3} + 2} \times 3 {x}^{2} = \\ = 5 ln( {x}^{3} + 2) + \frac{15 {x}^{3} }{ {x}^{3} + 2}$

$y' = {( {x}^{3} + 2)}^{5x} (5 ln( {x}^{3} + 2) + \frac{15 {x}^{3} }{ {x}^{3} + 2 } ) \\$

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