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Apr 12, 2019 at 10:48 comment added Michael Bächtold @user2520938 $D$(const.)=0 holds always. It follows from the Leibniz rule of derivations and $\mathbb{k}$-linearity.
Apr 12, 2019 at 8:41 comment added Michael Bächtold @user2520938 because any function $g$ can be decomposed as $g=g(x) + (g-g(x))$. The first summand is constant hence vanishes when applying a derivation, the second summand is a function with value zero at $x$.
Apr 12, 2019 at 6:52 comment added user2520938 why does it suffice to prove this for $f=0$?
Apr 11, 2019 at 21:34 vote accept Qfwfq
Apr 11, 2019 at 21:11 history answered Tom Goodwillie CC BY-SA 4.0