Question

cum verific f'(×)din asta? f(×)=(×+2)^3

Answer 1

Aplici regula ridicarii la putere si obtii:

$f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}(x+2)^3=3(x+2)^2\cdot\frac{\mathrm{d}}{\mathrm{d}x}(x+2)$

Cum parametrul $x$ are derivata 1 si constanta $2$ are derivata $0$:

$f'(x)=3(x+2)^2=3(x^2+4x+4)=3x^2+12x+12\:\forall\: x\in \mathbb{R}$

Answer 2

f(x)=(x+2)³

f(x)'=3(x+2)²

(a+b)²=a²+2ab+b²

f(x)'=3(x+2)²=3(x²+4x+4)=3x²+12x+12

f(x)'=3x²+12x+12