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Question

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

Solution

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Differentiating the both sides w.r.t. $x$

\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 $f(x)$ or any of its derivatives, the result is of the same form as the function to be differentiated by setting

$a_{n+1}=a_n$, $b_{n+1}=(2a_n+b_n)$ and $c_{n+1}=b_n+c_n$

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