Evaluate the following functions in differential operator form.
"1.\\ D\\left(x^{2}+2 x+3\\right)"
Since "D=\\frac{d}{d x}"
So, we have,
"\\begin{aligned}\n\n&\\frac{d}{d x}\\left(x^{2}+2 x+3\\right) \\\\\n\n&=\\frac{d}{d x}\\left(x^{2}\\right)+\\frac{d}{d x}(2 x)+\\frac{d}{d x}(3) \\\\\n\n&=2 x+2 \\\\\n\n&2.\\ D^{2}\\left(x e^{3 x}-e^{4 x}\\right)\n\n\\end{aligned}"
Since "D=\\frac{d}{d x}"
So, we have,
"\\begin{aligned}\n\n&\\frac{d^{2}}{d x^{2}}\\left(x e^{3 x}-e^{4 x}\\right) \\\\\n\n&\\frac{d}{d x}\\left(\\frac{d}{d x}\\left(x e^{3 x}-e^{4 x}\\right)\\right)\n\n\\end{aligned}"
"=\\frac{d}{d x}\\left(\\frac{d}{d x} x e^{3 x}-\\frac{d}{d x} e^{4 x}\\right)"
Recall the product rule of differentiation
"U V=V d u+U d v"
By comparison let "u=x, v=e^{3 x}"
Upon differentiation. We have,
"\\begin{aligned}\n\n&\\frac{d}{d x}\\left(e^{3 x}+3 x e^{3 x}-4 e^{4 x}\\right) \\\\\n\n&=\\frac{d}{d x}\\left(e^{3 x}\\right)+3 \\frac{d}{d x}\\left(x e^{3 x}\\right)-4 \\frac{d}{d x}\\left(e^{4 x}\\right) \\\\\n\n&=3 e^{3 x}+3\\left(e^{3 x}+3 x e^{3 x}\\right)-4\\left(4 e^{4 x}\\right) \\\\\n\n&=3 e^{3 x}+3 e^{3 x}+9 x e^{3 x}-16 e^{4 x} \\\\\n\n&=9 x e^{3 x}+6 e^{3 x}-16 e^{4 x}\n\n\\end{aligned}"
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