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¡Hola Joan!
Son muchos ejercicios, haré dos y si quieres manda los otros dos en otra pregunta.
$$\begin{align}&a) \frac{d^3}{dx^3}\left(e^{x^2-3x} \right)\\&\\&\frac{d}{dx}\left(e^{x^2-3x} \right)=(2x-3)\left(e^{x^2-3x} \right)\\&\\&\frac{d^2}{dx^2}\left(e^{x^2-3x} \right)=2\left(e^{x^2-3x} \right)+(2x-3)(2x-3)\left(e^{x^2-3x} \right)=\\&\\&\qquad(2+4x^2-12x+9)\left(e^{x^2-3x} \right)=(4x^2-12x+11)\left(e^{x^2-3x} \right)\\&\\&\frac{d^3}{dx^3}\left(e^{x^2-3x} \right)=(8x-12)\left(e^{x^2-3x} \right)+(4x^2-12x+11)(2x-3)\left(e^{x^2-3x} \right)=\\&\\&\qquad (8x-12+8x^3-12x^2-24x^2+36x+22x-33)\left(e^{x^2-3x} \right)=\\&\\&\qquad (8x^3-36x^2+66x-45)\left(e^{x^2-3x} \right)\\&\\&\\&\\&\\&b) \frac{d^3}{dx^3}(x^5e^{3x})\\&\\&\frac{d}{dx}(x^5e^{3x})=5x^4e^{3x}+x^5·3e^{3x}=(3x^5+5x^4)e^{3x}\\&\\&\frac{d^2}{dx^2}(x^5e^{3x})=(15x^4+20x^3)e^{3x}+3(3x^5+5x^4)e^{3x}=\\&\\&\qquad (15x^4+20x^3+9x^5+15x^4)e^{3x}=\\&\\&\qquad (9x^5+30x^4+20x^3)e^{3x}\\&\\&\frac{d^3}{dx^3}(x^5e^{3x})=(45x^4+120x^3+60x^2)e^{3x}+3(9x^5+30x^4+20x^3)e^{3x}=\\&\\&\qquad (45x^4+120x^3+60x^2+27x^5+90x^4+60x^3)e^{3x}=\\&\\&\qquad(27x^5+135x^4+180x^3+60x^2)e^{3x}\end{align}$$
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