Derivative of Identity Function

Theorem
Let $I_\R: \R \to \R$ be the identity function.

Then $\forall x \in \R: I_\R' \left({x}\right) = 1$.

Note that this can be more compactly written $D_x \left({x}\right) = 1$.

Proof for Real Numbers
The identity function is defined as $\forall x \in \R: I_\R \left({x}\right) = x$.

Thus:

Proof for Complex Numbers
The identity function is defined as $\forall x \in \C: I_\C \left({z}\right) = z$.

Thus:

Style Note
Using Leibniz's notation for derivatives $\left (\dfrac{\mathrm dy}{\mathrm dx}\right )$ this theorem can be stated as


 * $\displaystyle \frac{\mathrm dx}{\mathrm dx} = 1$

Which is not to say that derivatives are fractions, but the theorem is quite elegant this way.