Soit f(x)=(2x+1)(x2−3)f(x) = (2x+1)(x^2-3)f(x)=(2x+1)(x2−3). Quelle est f′(x)f'(x)f′(x) ?
4x4x4x
2(x2−3)+(2x+1)⋅2x2(x^2-3) + (2x+1) \cdot 2x2(x2−3)+(2x+1)⋅2x
2⋅2x2 \cdot 2x2⋅2x
(2x+1)(x2−3)(2+2x)(2x+1)(x^2-3)(2+2x)(2x+1)(x2−3)(2+2x)
Score: 0/0