I was reading a text where the transpose $A^T: V^\ast \to V^\ast$ of a linear operator $A: V\to V$ on vector space $V$ was defined as follows $(A^Tf)(v) = f(Av)$ with $f \in V^\ast, v \in V$. (So not the adjoint, no inner product was introduced yet). This was then used to illustrate that the derivative of the Dirac delta linear functional gives $$\left( \frac{d^T}{dx} \delta \right)[f(x)] = \delta[\frac{df(x)}{dx}] = f'(0).$$
How does this square with the usual $\delta' [f(x)] = -f'(0)$ in analogy to general Schwartz distributions (where the minus sign arises from the partial integration analogy)?