Răspuns:
a) f `(x)=(x+2) `*eˣ+(x+2)*eˣ`=eˣ+(x+2)eˣ=eˣ(1+x+2)=eˣ(x+3)
b) f `(x)= (3/x) `+lm `x +1 `= -3/x²+1/x+0 =-3/x²+1/x
c) f `(x)=[(x+3 )`(x²+2)-(x+3)(x²+2) `]/(x²+2)²=
(x²+2-2x(x+3)/(x²+2)²=(x²+2-2x²-6x)/(x²+2)²=
(-x²-6x+2)/(x²+2)²
Explicație pas cu pas: