(Производная)Очень нужна ваша помощь. Created by daryasvinar. algebra-ru

(Производная)Очень нужна ваша помощь...

daryasvinar @daryasvinar

July 2022 1 1 Report

(Производная)Очень нужна ваша помощь

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Universalka

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4) y ' = (2x³)' - (24x)' + 5' = 6x² - 24 + 0 = 6x² - 24
5) y ' = (2x³)' + (3x²)' - (12x)' + 5' = 6x² + 6x - 12
6) y ' = [(x + 2)²]' * (3x - 1) + (x + 2)² * (3x - 1)' = 2(x+2)(x+2)'(3x-1) +
+ 3(x + 2)²= (2x + 4)(3x - 1) + 3(x² + 4x + 4) = 6x² - 2x +12x - 4 + 3x² +12x +12=
=9x² +22x + 8
7) y ' = (x⁴)' - (4x³)' + (4x²)' = 4x³ - 12x² + 8x
14) $y' = \frac{(2x)'*( x^{2} +9)-2x*( x^{2} +9)'}{( x^{2} +9) ^{2} } = \frac{2( x^{2} +9)-2x*2x}{( x^{2} +9) ^{2} } = \frac{2 x^{2} +18-4 x^{2} }{( x^{2} +9) ^{2} } = \frac{18-2 x^{2} }{( x^{2} +9) ^{2} }$
15) $y' = \frac{(6x-6)'*( x^{2} +3)-(6x-6)*( x^{2} +3)'}{( x^{2} +3) ^{2} } = \frac{6( x^{2} +3)-2x(6x-6)}{( x^{2} +3) ^{2} } = \frac{6 x^{2} +18-12 x^{2} +12x}{( x^{2} +3) ^{2} }$ $= \frac{18+12x-6 x^{2} }{( x^{2} +3) ^{2} }$
16) $y' = \frac{( x^{2} -2x+2)'*(x-1)-( x^{2} -2x+2)*(x-1)'}{(x-1) ^{2} } = \frac{(2x-2)(x-1)-( x^{2} -2x+2)}{(x-1) ^{2} }=$ $\frac{2 x^{2} -2x-2x+2- x^{2} +2x-2}{(x-1) ^{2} }= \frac{ x^{2} -2x}{(x-1) ^{2} }= \frac{x(x-2)}{(x-1) ^{2} }$
17) $y' = (x-1)'* \sqrt{x} +(x-1)*( \sqrt{x} )' = \sqrt{x} +(x-1)* \frac{1}{2 \sqrt{x} } = \sqrt{x} + \frac{x-1}{2 \sqrt{x} }$ $= \frac{2x+x-1}{2 \sqrt{x} }= \frac{3x-1}{2 \sqrt{x} }$

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daryasvinar Спасибо большое!

25 ballovpodskazhitelyudi dobrye reshenie

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