Răspuns:

Explicație pas cu pas:

f(x)=(x²+x+1)eˣ

f `(x)=(2x+1)eˣ+(x²+x+1)eˣ=

eˣ(2x+1+x²+x+1)=

eˣ(x²+3x+2)

[tex]\boldsymbol{(f \cdot g)' = f' \cdot g + f \cdot g'}[/tex]

[tex]f'(x) = \Big((x^{2} +x+1)\Big)' = \Big(x^{2} +x+1\Big)' \cdot e^{x} + (e^{x})' \cdot (x^{2} +x+1) = \\[/tex]

[tex]= (2x + 1) \cdot e^{x} + e^{x} \cdot (x^{2} +x+1) = \boldsymbol{e^{x} \cdot (x^{2} + 3x + 2)}\\[/tex]

unde:

[tex]\Big(x^{2} +x+1\Big)' = (x^{2})' + x' + 1' = 2x + 1 + 0 = 2x + 1\\[/tex]

[tex](e^{x})' = e^{x}[/tex]