teorema rantai 1. f(x) = [tex] \sqrt{2 x^{2}+4x } [/tex] =

Dalil Rantai

f(x)= √(2x²+4x) = (2x² +4x)¹/²
.
u = 2x² + 4x
du/dx  = 4x + 4 = 4(x +1)
.
y = u¹/²
dy/du = 1/2  u⁻¹/²  = 1/ (2√u)

y' = dy/dx
dy/dx = dy/du . du/dx
y ' = 1/(2√u) . 4(x+1)

y' = 4(x+1) / 2√(2x²+4x)
y' = 2(x+1)/ √(2x²+4x)
atau
y' = {2(x+1)√(2x²+4x)} / (2x² +4x)
y' = 2(x+1)√(2x²+4x) / 2(x²+4x)
y' = (x+1)√(2x²+4x) /  (x²+4x)