Tentukan turunan pertama terhadap x y =.... Question from @mayanghejo

September 2018 2 59 Report

Tentukan turunan pertama terhadap x
y = $\frac{ (x^{2}+2x)^3 }{x^{3}-4x}$

arikamachroni U=(x²+2x)³ v=x³-4x=x(x²-4)=x(x+2)(x-2)=(x-2)(x²+2x)
u' = 3. (2x+2).(x²+2x)² v' = 3x²-4
= 6(x+1).(x²+2x)² v² = (x-2)².(x²+2x)²

y' =[u'.v - v'.u ]/ v² = [6(x+1)(x²+2x)³.(x-2)(x²+2x) - (3x² - 4)(x²+2x)³]/ [x³-4x]²

y' =(x²+2x)³ [6(x+1)(x-2) - 3x²+4] / (x²+2x)²(x-2)²

y' = (x²+2x) [ 6x² - 6x -12 - 3x² + 4] / (x-2)²

y' = x(x+2) [3x² - 6x - 8] / (x-2)²

y' = (x²+2x)(3x²-6x-8)/ (x-2)²

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Takamori37 $\displaystyle y=\frac{(x^2+2x)^3}{x^3-4x}=\frac{x^3(x+2)^3}{x(x+2)(x-2)}=\frac{x^2(x+2)^2}{x-2}=\frac{(x^2+2x)^2}{x-2} \\ g(x)=(x^2+2x)^2~~~\to~~~g'(x)=2(x^2+2x)(2x+2) \\ h(x)=x-2~~~~~~~~~\to~~~h'(x)=1 \\ y'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)} \\\\ y'=\frac{2(x-2)(x^2+2x)(2x+2)-(x^2+2x)^2}{(x-2)^2} \\\\ y'=\frac{(x^2+2x)(4(x-2)(x+1)-(x^2+2x))}{(x-2)^2} \\\\$

$\displaystyle y'=\frac{(x^2+2x)(4(x^2-x-2)-x^2-2x)}{(x-2)^2} \\\\ y'=\frac{x(x+2)(4x^2-4x-8-x^2-2x)}{(x-2)^2} \\\\ y'=\frac{x(x+2)(3x^2-6x-8)}{(x-2)^2}=\frac{x(3x^3-6x^2-8x+6x^2-12x-16)}{x^2-4x+4} \\\\ y'=\frac{x(3x^3-20x-16)}{x^2-4x+4} \\\\ y'=\frac{3x^4-20x^2-16x}{x^2-4x+4}$

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