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 $\map {f_c} z$ be the constant function on an open domain $D \in \C$, where $c \in \C$.


Then:

$\forall z \in D : \map { {f_c}'} z = 0$


Proof

The function $f_c: \R \to \R$ is defined as:

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

Thus:

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

$\blacksquare$


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

Retrieved from "https://proofwiki.org/w/index.php?title=Derivative_of_Constant&oldid=618474"