Cari turunan pertama fungsi f(x) = (2x + 1)^2 dengan caranya

f(x) = (2x+1)²
f(x) = (2x+1)(2x+1)

u = 2x+1
v = 2x+1

f'(x) = uv' + u'v
f'(x) = (2x+1)2 + 2(2x+1)
f'(x) = 4x+2+4x+2
f'(x) = 8x + 4
f'(x) = 4(2x+1)

Rumus:
Gunakan teorema rantai ↓↓
⇒ dy/dx = dy/du × du/dv × dv/dx

Misalkan: u = (2x + 1)
Maka: f(x) = u²

dy/du = 2u²⁻¹
dy/du = 2u

du/dx = 2

dy/dx = dy/du × du/dx
dy/dx = 2u × 2
dy/dx = 4u
dy/dx = 4(2x + 1)