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

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?

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:

$\lim_{\delta x \to 0} \frac {\delta(\sin x)}{\delta x} = \frac {d(\sin x)}{dx} = \cos x.$

It appears the answer is no.

Care to elaborate?

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 $\displaystyle \lim_{\delta \to 0} \frac{\sin \delta}{\delta} = 1$.

$\sin \delta x \to \delta x$ 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 $\displaystyle \lim_{\delta \to 0} \frac{\sin \delta}{\delta} = 1$.

Clearly, those limiting behaviours only apply as $\delta x \to 0$. A proof for either result can be obtained without resort to calculus.

More elegant, no doubt, but the gain in rigour comes at loss of clarity.

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".

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.

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.