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One can create a formal definition which yeilds such a result that \begin{equation}  \frac{df(x)}{d\mathbb{I}}=\frac{f((1-\delta)x)-f(x)}{(1-\delta)-1} \frac{df(x)}{d\mathbb{I}}=\frac{f(x)-f((1-\delta)x)}{\delta}  \end{equation} 
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\begin{equation}  \frac{de^x}{d\mathbb{I}}=\frac{e^xe^{-\delta x}-e^x}{(1-\delta)-1}=\frac{e^x(1-\delta z)-e^x}{(1-\delta)-1} \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}