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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!

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    $\begingroup$ I think your left-hand side is better expressed as $(fg)'$ or $\frac{d}{dx}(fg)$. $\endgroup$ – Randall Jun 12 at 17:09
  • $\begingroup$ There is no connection in the way (I think) you are trying to see. $\endgroup$ – Anurag A Jun 12 at 17:09
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    $\begingroup$ @AnuragA Perfectly possible, just want to be sure :)! $\endgroup$ – PrincepsMaximus Jun 12 at 17:11
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    $\begingroup$ 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. $\endgroup$ – Dave L. Renfro Jun 12 at 17:37
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    $\begingroup$ 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 $\sinh$. $\endgroup$ – Pedro Jun 12 at 19:50
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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.

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