Răspuns:a
Explicație pas cu pas:
a) f `((x)=(2x³-x+3) `*3ˣ+(2x³-x+3)*(3ˣ) `=
(6x²-1)*3ˣ+(2x³-x+3)*3ˣln3=
3ˣ[6x²+(2x³-x+3)ln3]
g) f `(x)=(2x) `sinx+2x (sinx) `=
2sinx+2x*cosx
i) f `(x)=(x³-2x²-x-5)`*2ˣ+(x³-2x²-x-5)(2ˣ)`=
(3x²-4x-1)2ˣ+(x³-2x²-x-5)2ˣln2=
2ˣ[3x²-4x-1+(x³-2x²-x-5)ln2]
j)f `(x)=(x+1) `(sinx-cosx)+(x+1)(sinx-cosx) `=
sinx-cosx+(x+1)(cosx+sinx)
