# Does the product rule for differentiation has anything to do with $\sin( \alpha + \beta)$?

When learning about differentation, I came along the product rule: $$D(f \cdot g) = f \cdot Dg + g \cdot Df$$ I immediately thought of this rule from trigonometry: $$\sin( \alpha + \beta ) = \sin(\alpha) \cdot \cos(\beta) + \cos(\alpha) \cdot \sin(\beta)$$ Is there any relation between these two rules? How can this similarity be explained? Has it something to do with geometry?

For those who don't see the relation that I see: if $$\alpha$$ is the 'normal' function and $$\beta$$ is the derivative, then you 'get': $$\sin \cdot D\cos + \cos \cdot D\sin$$ And thus if we name sine $$f$$ and cosine $$g$$, we 'get': $$f \cdot Dg + g \cdot Df$$

Of course, I write 'get' but I know you can't do this like that, or can you? That's exactly my question: is there any relation between those two rules/equations stated the top? Or is this some weird thought in my brain? Thanks in advance!

• I think your left-hand side is better expressed as $(fg)'$ or $\frac{d}{dx}(fg)$. – Randall Jun 12 at 17:09
• There is no connection in the way (I think) you are trying to see. – Anurag A Jun 12 at 17:09
• @AnuragA Perfectly possible, just want to be sure :)! – PrincepsMaximus Jun 12 at 17:11
• Recall the basic relation between trig functions and exponential functions, namely $e^{ix} = \cos x + i\sin x,$ and the fact that $e^{\alpha + \beta} = e^{\alpha} \cdot e^{\beta}.$ You might be able to some kind of connection from this. – Dave L. Renfro Jun 12 at 17:37
• The original reasoning also works with $\sin$ replaced by $\sinh$. Therefore, any possible connection should be based on some property that is shared by $\sin$ and $\sinh$. – Pedro Jun 12 at 19:50

One connection is if you set $$\alpha=\beta=x$$, you get that $$\sin(2x)$$ is the derivative of $$\sin^2(x)$$, which is easy to verify.