Why a function that is $0$ in a neighborhood of a point must have a $0$ derivative at that point?

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The question was to show that $supp (D^{\alpha}f) \subset supp (f)$ if $f\in C_c^k(\Bbb R^n)$

My teacher's solution was, Let $x \in (supp f)^c \implies f(x)=0 \implies D^{\alpha}f(x)=0$ (since $(supp f)^c$ is open)

I didn't understand this last step, I mean even if $(supp f)^c$ is open, why does this fact give that $D^{\alpha}f(x)=0$ ??

N.B.: $supp f$ is the support of the function $f$

Maybe my question in other words is, if $f(a)=0$, what condition do we need so that we can deduce that $f'(a)=0$ ?

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The topological support is closed by definition.

As the compliment of the support is open, there must exist an open neighborhood $U\ni x$ such that $f(y)=0$ for all $y\in U$. Thus all partial derivatives of $f$ must be $0$ at $x$, else the very definition of the derivative as the instantaneous rate of change would imply that $f$ must be non-zero somewhere in $U$. Think of the partial derivatives as forming the approximating plane for differentiable functions.

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The compactness of the support of $f$ is a sufficient condition for your implication.

Hint: Look at the definition of the first derivative as a limit.