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for $a>0$. However, one could have introduced the parameter $a$ if it was not there, and set it to one at the end to generate the seemingly arbitrary step. This of course required knowledge of where $a$ would appear on the right hand side, this knowledge creates the information that a factor of $-1/2$ is added. However one can ask the question, is it possible to define a derivative with respect to $1$. Such as to preserve the logic of \begin{equation}  \frac{d}{da}e^{ax}=xe^{ax} \\  \frac{d}{d1}e^{x}=xe^{x} \frac{d}{d\mathbb{I}}e^{x}=xe^{x}  \end{equation}