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\frac{de^x}{d\mathbb{I}}=\frac{e^x-e^xe^{-\delta x}}{\delta}=\frac{e^x-e^x(1-\delta x)}{\delta}=xe^x  \end{equation} We find as $e^x$ is the eigenfunction that produces an eigenvalue of $1$ with application of the $d/dx$ operator, $x$ is the eigenfunction that produces an eigenvalue of $1$ with application of the $d/d\mathbb{I}$ operator.