I want to evaluate the functional derivative of a function $f(x)$ with respect to its derivative $f'(x)$, i.e. $\delta f(x) / \delta f'(y)$. I have to results, and want to know if (1) the first one is correct, and (2) how to show it using the second.
Strategy 1: Expand $f(x)$ as an integral: $$ f(x) = f(0) + \int_0^x dz \; f'(z) \\ \frac{\delta f(x)}{\delta f'(y)} = \int_0^x dz \; \delta(z-y) = \Theta(y) - \Theta(y-x)$$
where $\Theta$ are step functions.
Question 1: Is this derivation correct?
Strategy 2: Expand $f(x)$ as a Taylor series: $$ f(x) = \sum_{l=0}^\infty \frac{x^l}{l!} \left ( \frac{d^l}{dz^l} f(z) \right ) \Bigg |_{z=0} = f(0) + \sum_{l=1}^\infty \frac{x^l}{l!} \left ( \frac{d^{l-1}}{dz^{l-1}} f'(z) \right ) \Bigg |_{z=0} $$
Now use the definition of the derivative of the Dirac delta: $$ \int dz \; g(z) \delta^{(l-1)} (z) = (-1)^{(l-1)} \int dz \; g^{(l-1)}(z) \delta(z) = (-1)^{(l-1)} g^{(l-1)}(0)$$
to write: $$ f(x) = f(0) + \sum_{l=1}^\infty \frac{x^l}{l!} (-1)^{(l-1)} \int dz \; f'(z) \delta^{(l-1)}(z) $$
Then: $$ \frac{\delta f(x)}{\delta f'(y)} = \sum_{l=1}^\infty \frac{x^l}{l!} (-1)^{(l-1)} \int dz \; \delta(x-z) \delta^{(l-1)}(z) $$
Now I don't know how to proceed.
Question 2: How do I show this is equivalent to the step function result above?
(Note: if I don't use the Dirac Delta derivative, then I can show the two are equivalent, since I can transform the Taylor series into the form $\int dz \; f'(z)$ in the first equation above).
Thanks!