proof of properties of derivatives by pure algebra
Theorem 1.
The derivative satisfies the following rules:
-
Linearity
ddx(f(x)+g(x))=dfdx+dgdx,ddx(af(x))=adfdx, for f(x),g(x)∈R[x] and a∈R.
-
Power Rule
ddx(xn)=nxn-1. -
Product Rule
ddx(f(x)g(x))=dfdxg(x)+f(x)dgdx.
Remark 2.
The following proofs apply to derivatives by pure algebra (http://planetmath.org/DerivativesByPureAlgebra). While the nature of the proofs are
similar to the usual proofs, the notion of a limit is replaced by modular
arithmetic in R[x,h]/(h).
Proof.
Power rule.
ddx(xn) | ≡ | (x+h)n-xnh | ||
= | n∑j=1(ij)xn-jhj-1 | |||
≡ | (n1)xn-1=nxn-1. |
Linearity rule. For all f(x),g(x)∈R[x]≅R[x,h]/(h), it follows
(f+g)(x+h)-(f+g)(x)h≡f(x+h)+g(x+h)-f(x)-g(x)h≡f(x+h)-f(x)h+g(x+h)-g(x)h. |
Furthermore, for all a∈R
(af)(x+h)-(af)(x)h≡af(x+h)-af(x)h=af(x+h)-f(x)h. |
Product rule. In R[x,h] modulo (h) we have:
ddx(fg) | ≡ | f(x+h)g(x+h))-f(x)g(x)h | ||
≡ | f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)h | |||
≡ | (f(x+h)-f(x))g(x+h)+f(x)(g(x+h)-g(x))h | |||
≡ | f(x+h)-f(x)hg(x+h)+f(x)g(x+h)-g(x)h | |||
≡ | dfdxg(x)+f(x)dgdx. |
∎
Title | proof of properties of derivatives by pure algebra |
---|---|
Canonical name | ProofOfPropertiesOfDerivativesByPureAlgebra |
Date of creation | 2013-03-22 16:00:03 |
Last modified on | 2013-03-22 16:00:03 |
Owner | Algeboy (12884) |
Last modified by | Algeboy (12884) |
Numerical id | 5 |
Author | Algeboy (12884) |
Entry type | Proof |
Classification | msc 26B05 |
Classification | msc 46G05 |
Classification | msc 26A24 |
Related topic | RulesOfCalculusForDerivativeOfPolynomial |