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

September 2018 2 57 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

1 votes Thanks 0

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

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

0 votes Thanks 0

About Us

Life Enjoy

" Life is not a problem to be solved but a reality to be experienced! "

Selesaikan persamaan integral

Pasal dalam UUD 1945 yang isinya pembentukan karakter bangsa?

Pendirian cagar alam di ujung kulon jawa barat bertujuan untuk...a. melestarikan badak bercula satub. mengembangbiakan tumbuhan langkac.melindungi kondisi alam ujung kulondua guru dua jawaban beda, antara a sama c. kira-kira yang mana ya? makasih sebelumnya :)

Recommend Questions

Life Enjoy

Get in touch

Social