An algebraic and general proof for the derivative of constant is 0

In this article, I wanna introduce an algebraic way to prove that the derivative of a constant function is 0.

Consider an $R-\mathrm{Algebra}:M$$R-\mathrm{Algebra}:M$ and an operator $\delta :M\to M$$\delta:M\to M$, $(M,\delta )$$(M,\delta)$

The delta is satisfied

Module homomorphism

$\delta (r_1{m}_1+r_2{m}_2)=r_1\delta (m_1)+r_2\delta (m_2)$$\delta(r_1m_1+r_2 m_2)=r_1\delta(m_1)+r_2\delta(m_2)$

Leibniz law

$\delta (ab)=\delta (a)b+a\delta (b)$$\delta(ab)=\delta(a)b+a\delta (b)$

So $\delta (1)=\delta (1\cdot 1)=1\delta (1)+\delta (1)1=\delta (1)+\delta (1)\Rightarrow \delta (1)=0$$\delta(1)=\delta(1\cdot 1)=1\delta(1)+\delta(1)1=\delta(1)+\delta(1)\Rightarrow\delta (1)=0$

And because of $\mathrm{\forall }r\in R,\delta (r)=r\delta (1)=r\cdot 0=0$$\forall r\in R,\delta(r)=r\delta(1)=r\cdot0=0$

And we know all the differentiable functions can be an $\mathbb{R}-\mathrm{Algebra}$$\mathbb R-\mathrm{Algebra}$

And $\begin{array}{c}\frac{d}{dx}\end{array}$$\frac{d}{dx}\\$ is a linear map that satisfied Leibniz law

Thus $\begin{array}{c}(C^1,\frac{d}{dx})\end{array}$$\left(C^1,\frac{d}{dx}\right)\\$ is a special condition for the $(M,\delta )$$(M,\delta)$

 

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