cara kyk diatas teorema rantai 1. f(x) = (x^2 - 2x)^5 = 2. f(x) = 1 / (x-5)^4 =

Teorema Rantai

1. f(x)= (x² -2x)⁵
u = x² - 2x  --> du/dx = 2x -2
y = u⁵ --> dy/du = 5 u⁴

y' = dy/dx
dy/dx = dy/du . du/dx
y ' = 5u⁴ . (2x - 2)
y' = 5(2x - 2)(x² -2x)⁴
y' = (10x -10)(x² -2x)⁴

2) f(x)= 1/(x-5)⁴
f(x)= (x-5)⁻⁴
u = x-5 --> du/dx  = 1
y = u⁻⁴ --> dy/du = - 4 u⁻⁵

y' = dy/dx
dy/dx = dy/du . du/dx
y ' = - 4 u⁻⁵ . 1
y ' = - 4 / (x -5)⁵

1. f(x) = (x^2 - 2x)^5
y = u^5
dy/du= 5u^4

u= x^2 - 2x
du/dx= 2x -2
f'(x) = dy/du . du/dx
= 5u^4 (2x - 2)
= (10x - 10) (x^2 - 2x)^4

2. f(x) = 1/(x-5)^4
= (x-5)^-4

y=u^-4
dy/du = -4u^-5

u= x - 5
du/dx = 1
f'(x) = dy/du . du/dx
= -4u^-5 (1)
=-4 (x-5)^-5

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