Derivative of Constant

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map {f_c} x$ be the constant function on $\R$, where $c \in \R$.


Then:

$\map {f_c'} x = 0$


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:

$\forall x \in \R: \map {f_c} x = c$

Thus:

\(\displaystyle \map {f_c'} x\) \(=\) \(\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map {f_c} {x + \delta x} - \map {f_c} x} {\delta x}\) 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 $\forall x: \map {f'} x = 0$.


Sources