# Math Help - connection derivation in algebra and calculus

1. ## connection derivation in algebra and calculus

in algebra, definition of derivation is:
let R be ring or nearing or primering etc..
an additive mapping d:R\toR said to be derivation on R if d(xy)=d(x)y+yd(x)
for all x,y\inR

when i come with example, generaly given the commutator f(x)= ax-xa. of course that satisfy aditive mapping and derivation.
I want to make different example, and confuse with derivation in calculus, cause let f(x)= 2x^2+4, of course f'(x)=4x but f(x) doesn't satisfy additive maping nor derivation
is there any connection between them?
(i thin'k derivation in algebra similar to product rule in calculus, but this make me more confused)

2. Originally Posted by wingz00
in algebra, definition of derivation is:
let R be ring or nearing or primering etc..
an additive mapping d:R\toR said to be derivation on R if d(xy)=d(x)y+yd(x)
for all x,y\inR

when i come with example, generaly given the commutator f(x)= ax-xa. of course that satisfy aditive mapping and derivation.
I want to make different example, and confuse with derivation in calculus, cause let f(x)= 2x^2+4, of course f'(x)=4x but f(x) doesn't satisfy additive maping nor derivation
is there any connection between them?
(I think derivation in algebra similar to product rule in calculus, but this make me more confused)
The statement that I have bolded is the key to this. The operation of differentiation satisfies the Leibniz rule for differentiating a product, and this is exactly the same as the defining property of a derivation.

For example, let R be the ring of all polynomials (over some field), and define D (a mapping from R to itself) by D(p(x)) = p'(x). Then D is an additive mapping on R, and D(p(x)*q(x)) = D(p(x))*q(x) + p(x)*D(q(x)). Thus D (differentiation) is a derivation.

3. Originally Posted by wingz00
in algebra, definition of derivation is:
let R be ring or nearing or primering etc..
an additive mapping d:R\toR said to be derivation on R if d(xy)=d(x)y+yd(x)
$=d[f(t)]=d(2t^2+4)=d(\underbrace{2t^2}_{x})+d(\underbrace{4}_{y})=d(2t^2)+d(4)=$
$d(2)t^2+2d(t^2)+d(4)=d(2)t^2+2[d(t)t+td(t)]+d(4)$