$f$ not underdetermined then does $df$ use only as many inputs as needed yet still equal that with more inputs than needed?

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I'm a physicist working in measure theory on a probability space and need to show that, essentially, the derivative of any measurable function, $f$, with a set of inputs (at most countable) equals $Z_{\Lambda}(f):=\int_\Lambda d\mu_x \frac{\delta}{\delta\mu_x} f$ such that the set of inputs, $\Lambda$, can be taken to be as small as possible and $\mu_x$ is a Wiener measure such that $x=\vec x(t=0)$. That is, I want to show that if $|\Lambda|=n$, $|\Lambda'|=n'$ and $n>n'$, then $Z_{\Lambda}=Z_{\Lambda'}$ but that this is only valid until the cardinality of the input set, $|\Lambda|$, reaches the minimum number of variables that $f$ can have.

My thoughts on a proof for this are towards representing $f$ as a matrix/more general tensor of equations such that the number of variables are determined by constraint counting so that, in order for $f$ to have a (unique) derivative, $f$ cannot be underdetermined. I think this will force a minimum in the cardinality of $f$ yet allow for the constraint on $Z_\Lambda$ that if $|\Lambda'|>|\Lambda|$ then $Z_{\Lambda}=Z_{\Lambda'}$. However, I would prefer a pre-existing (published/preprinted) proof if possible.

So, my question is: i) Is there a mathematical proof for this; and, if not, ii) does anyone have some ideas as to how I could improve the logic behind my outlined thoughts on a proof?