When was the derivative of sin(X) first determined ?

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seals2001
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I want to know in the timeline of mathematics , considering what was considered as far as had been discovered at that time , where I place ?

For example, if you can calculate a derivative from first principles you are at the same point Newton was when he discovered calculus.

Am I still at that point in time ? I assume so. So I have the same knowledge that the best mathematicians of the 17th century had !
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mqb2766
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(Original post by seals2001)
I want to know in the timeline of mathematics , considering what was considered as far as had been discovered at that time , where I place ?

For example, if you can calculate a derivative from first principles youu are at the same point Newton was when he discovered calculus.

Am I still at that point in time ? I assume so. So I have the same knowledge that the best mathematicians of the 17th century had !
Errr can you differentiate sin() from first principles?
Have a read of a good maths history book, Newton did a fair bit more than that and others had been kicking around ideas about calculus for 50-100 years before, but he unified the approach and applied it to understand gravity/elliptical motion, amongst other things.
Last edited by mqb2766; 3 weeks ago
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lordaxil
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(Original post by mqb2766)
Errr can you differentiate sin() from first principles?
Sure you can, using identity for \sin(A+B) = \sin A \cos B + \cos A \sin B:

\frac {\delta(\sin x)}{\delta x} = \frac {\sin(x+\delta x)-\sin x}{\delta x} = \frac {\sin x \cos(\delta x)+\cos x \sin(\delta x)-\sin x }{\delta x}.

Then, let \delta x\to 0 so \sin (\delta x)\to \delta x and \cos (\delta x)\to 1 and so:

\lim_{\delta x \to 0} \frac {\delta(\sin x)}{\delta x} = \frac {d(\sin x)}{dx} = \cos x.
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DFranklin
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(Original post by mqb2766)
Errr can you differentiate sin() from first principles?
(Original post by lordaxil)
Sure you can, using identity for \sin(A+B) = \sin A \cos B + \cos A \sin B:

\frac {\delta(\sin x)}{\delta x} = \frac {\sin(x+\delta x)-\sin x}{\delta x} = \frac {\sin x \cos(\delta x)+\cos x \sin(\delta x)-\sin x }{\delta x}.

Then, let \delta x\to 0 so \sin (\delta x)\to \delta x and \cos (\delta x)\to 1 and so:

\lim_{\delta x \to 0} \frac {\delta(\sin x)}{\delta x} = \frac {d(\sin x)}{dx} = \cos x.
It appears the answer is no.
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lordaxil
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(Original post by DFranklin)
It appears the answer is no.
Care to elaborate?
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DFranklin
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(Original post by lordaxil)
Care to elaborate?
\sin \delta x \to \delta x isn't a meaningfull statement when talking about limits. You can't pick and choose which bits of the limit you move outside. If I give you the benefit of the doubt about \sin \delta x \to \delta x, it's still a result claimed without proof. Same for \cos \delta x \to 1. In addition, \cos \delta x \to 1 isn't enough to deduce the result you want.

Edit: FWIW, it's a little less obvious but if you use the trig sum<->product formulas you can show \sin(x+\delta) - \sin(x) = 2 \sin(\delta/2) \cos(x+\delta/2), which reduces the "limit type" results you need to just \displaystyle \lim_{\delta \to 0} \frac{\sin \delta}{\delta} = 1.
Last edited by DFranklin; 3 weeks ago
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DFranklin
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(Original post by lordaxil)
Care to elaborate?
I also just realised you're not the OP. mqb2766's question was to point out that if the OP wants to compare themselves to Newton, there's a definite destinction between knowing the derivative of sin(x) and knowing how to *prove* that's the derivative. So the word "you" in that question was important.
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lordaxil
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(Original post by DFranklin)
\sin \delta x \to \delta x isn't a meaningfull statement when talking about limits. You can't pick and choose which bits of the limit you move outside. If I give you the benefit of the doubt about \sin \delta x \to \delta x, it's still a result claimed without proof. Same for \cos \delta x \to 1. In addition, \cos \delta x \to 1 isn't enough to deduce the result you want.
Clearly, those limiting behaviours only apply as \delta x \to 0. A proof for either result can be obtained without resort to calculus.

(Original post by DFranklin)
Edit: FWIW, it's a little less obvious but if you use the trig sum<->product formulas you can show \sin(x+\delta) - \sin(x) = 2 \sin(\delta/2) \cos(x+\delta/2), which reduces the "limit type" results you need to just \displaystyle \lim_{\delta \to 0} \frac{\sin \delta}{\delta} = 1.
More elegant, no doubt, but the gain in rigour comes at loss of clarity.
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seals2001
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(Original post by DFranklin)
\sin \delta x \to \delta x isn't a meaningfull statement when talking about limits. You can't pick and choose which bits of the limit you move outside. If I give you the benefit of the doubt about \sin \delta x \to \delta x, it's still a result claimed without proof. Same for \cos \delta x \to 1. In addition, \cos \delta x \to 1 isn't enough to deduce the result you want.

Edit: FWIW, it's a little less obvious but if you use the trig sum<->product formulas you can show \sin(x+\delta) - \sin(x) = 2 \sin(\delta/2) \cos(x+\delta/2), which reduces the "limit type" results you need to just \displaystyle \lim_{\delta \to 0} \frac{\sin \delta}{\delta} = 1.
Can you elaborate why that isn't a proof ?

Thanks.
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DFranklin
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(Original post by lordaxil)
Clearly, those limiting behaviours only apply as \delta x \to 0. A proof for either result can be obtained without resort to calculus.
I'm not nitpicking about whether you explicitly say "as \delta x \to 0"; it's clear enough from context. But my question is: what is \sin \delta x \to \delta x even supposed to mean? If you mean \displaystyle \lim_{\delta x \to 0} \frac{\sin \delta x}{\delta x} = 1, you should say so (and if you don't then I really don't know what you mean).

More fundamentally, you can't (always) look at a limiting expression and say "I'll take the limits for those bits out of the limit, but I'll leave some other bits in".

In your limit you've claimed that \displaystyle \lim_{\delta x \to 0} \frac{\sin x \cos \delta x - \sin x}{\delta x} = 0 by observing that \cos \delta x \to 1.

This is not a valid arguement. If we rearrange slightly as: \displaystyle \lim_{\delta x \to 0} \sin(x) \frac{\cos \delta x - 1}{\delta x} we see that you are dividing \cos \delta x - 1 by \delta x.

So the mere knowledge that \cos \delta x \to 1 is insufficient; you need instead \displaystyle \lim_{\delta x \to 0} \frac{\cos \delta x - 1}{\delta x} = 0, which is a much stronger statement..

More elegant, no doubt, but the gain in rigour comes at loss of clarity.
Not much point in a "proof" that's clear but wrong. Rearranging to avoid the need of results on how quickly \cos \delta x \to 1 is worth it, IMNSHO.
Last edited by DFranklin; 3 weeks ago
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lordaxil
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(Original post by DFranklin)
I'm not nitpicking about whether you explicitly say "as \delta x \to 0"; it's clear enough from context. But my question is: what is \sin \delta x \to \delta x even supposed to mean? If you mean \displaystyle \lim_{\delta x \to 0} \frac{\sin \delta x}{\delta x} = 1, you should say so (and if you don't then I really don't know what you mean).
Yes, that's what I mean. As you say, it couldn't really mean anything else.

(Original post by DFranklin)
More fundamentally, you can't (always) look at a limiting expression and say "I'll take the limits for those bits out of the limit, but I'll leave some other bits in".
Fair point. In this case, it is OK to do though.

(Original post by DFranklin)
Not much point in a "proof" that's clear but wrong. Rearranging to avoid the need of results on how quickly \cos \delta x \to 1 is worth it, IMNSHO.
I wasn't aiming for full mathematical rigour - just to show that sin(x) can be differentiated from first principles, which you have agreed with me is possible.
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DFranklin
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(Original post by lordaxil)
Yes, that's what I mean. As you say, it couldn't really mean anything else.



Fair point. In this case, it is OK to do though.
No, it isn't.

I wasn't aiming for full mathematical rigour - just to show that sin(x) can be differentiated from first principles, which you have agreed with me is possible.
Yes it can be done. But you didn't do so.
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lordaxil
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Yes, you are right. What I wrote originally was wrong and I see that now. Correction repped.
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