The sin(a+b) formula and the product rule for derivatives

[kotreny]

Why are they similar?

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

d/dx (f(x)*g(x))=f(x)g'(x)+g(x)f'(x)

Somewhere on this very site there was mention of this, I believe, though I can't remember where. Maybe I'm delirious.

 

[Gib Z]

I don't see how they are very similar actually? What do you mean by "similar" ?

I think perhaps what you saw was that the formula for repeated differentiation of products was "similar" to the binomial theorem.

 

[kotreny]

Admittedly, the word "similar" might have been too strong to describe the pattern I see. Just the vague shapes of the formulas are the same; everything else is quite different.

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

The two addends each contain two factors. The same functions are used in both factors (sin, cos), although the "subjects" alternate (a, b forgive the terminology.)

d/dx (f(x)*g(x))=f(x)g'(x)+g(x)f'(x)

In this case the functions are :nothing: and :prime:. The subjects are f(x) and g(x).

Yes, a very general pattern, but I feel it is like the similarity between a frog and a giraffe: They are both animals, with a heart, digestive system, etc. Maybe some basic property of arithmetic applied at the very beginning of the derivations for these somehow did this. Maybe it has something to do with the binomial theorem. There might be other formulas following this pattern too, in which case I'd be happy to know them. I'm looking for a broad, underlying principle.

 

[sokrates]

Why are they similar?

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

d/dx (f(x)*g(x))=f(x)g'(x)+g(x)f'(x)

Somewhere on this very site there was mention of this, I believe, though I can't remember where. Maybe I'm delirious.

Let's see for a simple case: If you call $$sin(x) = f(x)$$ and also $$sin(x) = g (x)$$

$$sin(x+x)=sin(x)cos(x) + cos(x)sin(x)$$ is really like
$$\quad \quad \quad = f(x)g'(x) + g(x)f'(x)$$

and this is equal to $$d/dx (f(x) g(x) )$$ as you said... But that makes sense, since

$$\frac{d}{dx} sin^2 (x) = 2 sin(x) cos(x) = sin(2x) = sin (x+x)$$

Apart from these simple identities, I don't see a fundamental reason for this similarity.