Well, I am not exactly an expert on the notation. Like I said, it's usually only used in rare cases, and when it is used, it's usually clear what it means. The example you posted is somewhat unusual because, as far as I can tell, it's not even valid notation. I stated in my last post what I believe was intended by the notation, but I do not accept that the notation as written is valid. But again, I'd yield to someone who is more familiar with it and can explain why the notation is valid.

\(\displaystyle D\) is an operator that maps a function to another function (its derivative). If I am given a function \(\displaystyle f\), then \(\displaystyle Df\) is the derivative of \(\displaystyle f\). That is, \(\displaystyle Df\) is itself a function. This isn't a "product" because we aren't multiplying \(\displaystyle D\) with \(\displaystyle f\).

So you can evaluate that function at a value \(\displaystyle x\), i.e., \(\displaystyle (Df)(x)\). The extra parentheses are a bit confusing, but are necessary to clarify that we want to evaluate the function \(\displaystyle Df\) at \(\displaystyle x\). If we wrote \(\displaystyle Df(x)\) it may appear we are multiplying \(\displaystyle f(x)\) by some number \(\displaystyle D\).

So, the left-hand side of the notation you posted really means evaluate the function \(\displaystyle De^\gamma\) at \(\displaystyle x\). Which means \(\displaystyle e^\gamma\) should be a function. The problem is that we don't really know what \(\displaystyle e^\gamma\) means without guessing.

I'm being pedantic here, and I think this is probably a minor point in whatever you are working on. So, I might suggest ignoring what you posted, and instead consider this example instead:

Suppose \(\displaystyle f\) is a function such that \(\displaystyle f(x)=e^{\gamma x}\). Then \(\displaystyle (Df)(x) = \gamma e^{\gamma x}\).