Derivative of Constant

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Theorem

Let $f_c \left({x}\right)$ be the constant function on $\R$, where $c \in \R$.


Then:

$f_c' \left({x}\right) = 0$


Complex Domain

Let $f_c \left({z}\right)$ be the constant function on an open domain $D \in \C$, where $c \in \C$.


Then:

$\forall z \in D : f_c' \left({z}\right) = 0$


Proof

The function $f_c: \R \to \R$ is defined as $\forall x \in \R: f_c \left({x}\right) = c$.

Thus:

\(\displaystyle f_c' \left({x}\right)\) \(=\) \(\displaystyle \lim_{\delta x \mathop \to 0} \frac {f_c \left({x + \delta x}\right) - f_c \left({x}\right)} {\delta x}\) Definition of Differentiation
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\delta x \mathop \to 0} \frac {c - c} {\delta x}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\delta x \mathop \to 0} \frac 0 {\delta x}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

$\blacksquare$


Also see

This is the converse of Zero Derivative implies Constant Function.

Thus we see that $f$ is the constant function if and only if $\forall x: f' \left({x}\right) = 0$.


Sources