Răspuns:
f;(0,+∞)→R f(x)=(x-2)lnx
f `(x)=(x-2) `lnx+(x-2)*(lnx)`=lnx+(x-2)/x
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b. f(x)=(x²+2x+3)eˣ
f `(x)=(x²+2x+3) `*eˣ+(x²+2x+3)(eˣ)`=
(2x+2)eˣ+(x²+2x+3)eˣ=
eˣ(2x+2+x²+2x+3)=
eˣ(x²+4x+5)
Laa pct b in paranteza intre x² si 2x cred ca este + nu ×
Explicație pas cu pas:
