proof of properties of derivatives by pure algebra

Theorem 1.

The derivative satisfies the following rules:

• $$\frac{d}{dx}(f(x)+g(x))=\frac{df}{dx}+\frac{dg}{dx},\frac{d}{dx}(af(x))=a\frac{df}{dx},$$

  for $f(x),g(x)\in R[x]$ and $a\in R$.

• $$\frac{d}{dx}(x^n)=n{x}^{n-1}.$$

• $$\frac{d}{dx}(f(x)g(x))=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}.$$

Remark 2.

The following proofs apply to derivatives by pure algebra (http://planetmath.org/DerivativesByPureAlgebra). While the nature of the proofs aresimilar to the usual proofs, the notion of a limit is replaced by modulararithmetic in $R\mathit{}\mathrm{[}x\mathrm{,}h\mathrm{]}\mathrm{/}\mathrm{(}h\mathrm{)}$.

Proof.

Power rule.

${\displaystyle \frac{d}{dx}}(x^n)$$\equiv$${\displaystyle \frac{{(x+h)}^n-x^n}{h}}$

$=$${\displaystyle \sum _{j=1}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)x^{n-j}{h}^{j-1}$

$\equiv$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{1}}\right)x^{n-1}=n{x}^{n-1}.$

Linearity rule.For all $f(x),g(x)\in R[x]\cong R[x,h]/(h)$, it follows

$$\frac{(f+g)(x+h)-(f+g)(x)}{h}\equiv \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\equiv \frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}.$$

Furthermore, for all $a\in R$

$$\frac{(af)(x+h)-(af)(x)}{h}\equiv \frac{af(x+h)-af(x)}{h}=a\frac{f(x+h)-f(x)}{h}.$$

Product rule.In $R[x,h]$ modulo $(h)$ we have:

${\displaystyle \frac{d}{dx}}(fg)$$\equiv$${\displaystyle \frac{f(x+h)g(x+h))-f(x)g(x)}{h}}$

$\equiv$${\displaystyle \frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}}$

$\equiv$${\displaystyle \frac{(f(x+h)-f(x))g(x+h)+f(x)(g(x+h)-g(x))}{h}}$

$\equiv$${\displaystyle \frac{f(x+h)-f(x)}{h}}g(x+h)+f(x){\displaystyle \frac{g(x+h)-g(x)}{h}}$

$\equiv$${\displaystyle \frac{df}{dx}}g(x)+f(x){\displaystyle \frac{dg}{dx}}.$

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