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Show that the nth derivative of (ax2+bx+c)ex\left(a x^{2}+b x+c\right) e^{x} is a function of the same form but with different constants.

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Answered 7 months ago
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Differentiating the both sides w.r.t. xx

f(x)=(ex)(ax2+bx+c)+eax(ax2+bx+c)f(x)=ex(ax2+bx+c)+eax(2ax+b)f(x)=ex(ax2+bx+c)+ex(2ax+b)f(x)=ex(ax2+(2a+b)x+b+c)f(x)=ex(ax2+bx+c)\begin{align*} &f'(x)=\left(e^{x}\right)'(ax^2+bx+c)+e^{ax}(ax^2+bx+c)'\tag{Product rule}\\ \Rightarrow &f'(x)=e^{x}(ax^2+bx+c)+e^{ax}(2ax+b)\tag{general power rule, exponential differentiation and chain rule}\\ \Rightarrow &f'(x)=e^{x}(ax^2+bx+c)+e^{x}(2ax+b)\\ \Rightarrow &f'(x)=e^{x}(ax^2+(2a+b)x+b+c)\\ \Rightarrow &f'(x)=e^{x}(a^{\prime}x^2+b^{\prime}x+c^{\prime})\tag{Set $a^{\prime}=a$, $b^{\prime}=(2a+b)$ and $c^{\prime}=b+c$} \end{align*}

Hence every time we differentiate f(x)f(x) or any of its derivatives, the result is of the same form as the function to be differentiated by setting

an+1=ana_{n+1}=a_n, bn+1=(2an+bn)b_{n+1}=(2a_n+b_n) and cn+1=bn+cnc_{n+1}=b_n+c_n

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