Differentiating trig by first principles

Please see photo where I prove that sinx differentiates to cosx.
Why do we use the fact that as dx approaches 0, sindx approaches dx, but we ignore that as dx approaches 0, sindx approaches 0?
Surely these two points are equally significant and should both be included in this proof?


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A brief comment:
It's more conventional to consider small $\delta x$ first, and then consider the limit as $\delta x \rightarrow 0$ which is then written $dx$.
The reason for this is that it helps clear up the confusion, as you can see, both the numerator and the denominator have a $\delta x$ term, which tend to zero at the same rate, and we see that:
$\cos(x) \dfrac{\delta x}{\delta x} =\cos(x) \cdot 1=\cos(x)$.

If you had a $(\delta x)^2$ term in the numerator, as can appear in more complicated proofs, then you would be right in worrying about the additional factor:
$\cos(x) \dfrac{(\delta x)^2}{\delta x} =\cos(x) \cdot \delta x \rightarrow 0$ as $\delta x \rightarrow 0$.

By dx I meant small delta X I just don't know how to write it properly. I'm still a bit confused though with regards to my original question.


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Essentially, you write $\sin(\delta x) \approx \delta x$ for small $\delta x$, you worry about taking the limit as $\delta x \rightarrow 0$ after writing $\dfrac{\delta x}{\delta x}=1$.
Clearly, $\lim_{\delta x \rightarrow 0}1=1$.

I see what you mean but I don't feel you're explaining why this is the case. Sorry to be annoying!


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Please see photo where I prove that sinx differentiates to cosx.
Why do we use the fact that as dx approaches 0, sindx approaches dx, but we ignore that as dx approaches 0, sindx approaches 0?
Surely these two points are equally significant and should both be included in this proof?


Posted from TSR Mobile

I'm sorry, but what you've written there is not a 'proof'. And "sin dx approaches dx" (or "sin (delta x) approaches delta x" is not a mathematical statement either.

Is this an argument you've come up with yourself, or is this following an example given in a book somewhere?

Sorry to be picky - it's right to want to understand how a proof works, but you absolutely have to start with something that is properly constructed in the first place!

I'm sorry, but what you've written there is not a 'proof'. And "sin dx approaches dx" (or "sin (delta x) approaches delta x") is not a mathematical statement either.

Is this an argument you've come up with yourself, or is this following an example given in a book somewhere?

Sorry to be picky - it's right to want to understand how a proof works, but you absolutely have to start with something that is properly constructed in the first place!

Can you see that:
$\dfrac{\sin(\delta x)}{\delta x} \approx \dfrac{\delta x}{\delta x} = 1$?

I think this is an A-level "proof" I seem to recall reading something virtually identical in my textbook a few years ago.
I do, however, agree the write up could do with some work.

Yes of course


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Are you guys all at uni now or what? I'm interested to know. Sorry if I sound stupid...I promise I'm not....I am a scholar and have already 3 offers to study maths. I just find it hard to understand things I haven't been taught...as anyone would.


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My maths teacher told me this was correct but obviously it is not then....can you tell me what's wrong with it because just saying it is wrong isn't very helpful. Also I just called it a proof because I didn't know what else to call it...let's not be too OTT now


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I think this is an A-level "proof" I seem to recall reading something virtually identical in my textbook a few years ago.
I do, however, agree the write up could do with some work.

Then surely: $\cos(x)\dfrac{\sin(\delta x)}{\delta x} \approx \cos(x) \dfrac{\delta x}{\delta x}=\cos(x)$.
With a little hand-waving,you can slip in an equals sign because the smaller $\delta x$ gets the better the approximation above becomes.

In the limit as $\delta x \rightarrow 0$ it can be shown that, indeed, $\dfrac{\sin(\delta x)}{\delta x} \rightarrow 1$.
(There are a variety of ways to do this, though the ones I know are beyond the scope of an A-Level course, they shouldn't be all that hard to follow. If you're interested, a quick google search will provide some examples, the easiest method is probably to use L'Hôpital.)



Yeah, I'm a second year mathmo. Davros is more experienced than me, especially at explaining stuff. (I'm reasonably sure he's a teacher - provided memory serves).

In the limit as $\delta x \rightarrow 0$ it can be shown that, indeed, $\dfrac{\sin(\delta x)}{\delta x} \rightarrow 1$.
(There are a variety of ways to do this, though the ones I know are beyond the scope of an A-Level course, they shouldn't be all that hard to follow. If you're interested, a quick google search will provide some examples, the easiest method is probably to use L'Hôpital.)



Yeah, I'm a second year mathmo. Davros is more experienced than me, especially at explaining stuff. (I'm reasonably sure he's a teacher - provided memory serves).

Sorry if I was sounding harsh - I appreciate that you are an A level student and this must be frustrating. If you'd said something like "is this OK as a plausible argument why the derivative of sin x is cos x?" then I would probably not have commented. My real concern is whether something like this is being presented as a 'proof' in a textbook, because it's really not precise enough to be called a proof




The problem is really with the innocent-looking statement "$\sin h \approx h$ when h is small" (I'm using h instead of delta x just to simplify my typing!). We all tend to look at this and say "Oh yes, sin h is approximately equal to h", but what does that actually mean mathematically? For example, it's true that $x^2 \approx x$ when x is close to 0 because both quantities are close to 0, but that doesn't mean we can deduce that $x^2 / x \to 1$ as x->0 - we know that the limit is in fact 0.

What we need here is actually that (sin h)/h -> 1 as h->0 and this is in fact the case, provided that h is measured in radians.

For info, here's what I regard as the 'standard' A level proof that the OP needs - taken from the 'bible' Bostock & Chandler. I assumed it was included in all standard A level texts,
but perhaps I'm just terribly naive

Please, please don't suggest l'Hopital here! l'Hopital would be an example of circular reasoning, because in order to evaluate the limit of (sin x) / x as x->0 you need to know the derivative of sin x at x = 0, and you can't calculate that without knowing what the limit is first!

The proof in that book does however skim over the evaluation of the limit:
$\displaystyle\lim_{\delta x \to 0} \dfrac{\sin(\frac{\delta x}{2})}{\frac{\delta x}{2}}$. In a way that I think the OP may have found dissatisfactory.

.

Sorry if I was sounding harsh - I appreciate that you are an A level student and this must be frustrating. If you'd said something like "is this OK as a plausible argument why the derivative of sin x is cos x?" then I would probably not have commented. My real concern is whether something like this is being presented as a 'proof' in a textbook, because it's really not precise enough to be called a proof




The problem is really with the innocent-looking statement "$\sin h \approx h$ when h is small" (I'm using h instead of delta x just to simplify my typing!). We all tend to look at this and say "Oh yes, sin h is approximately equal to h", but what does that actually mean mathematically? For example, it's true that $x^2 \approx x$ when x is close to 0 because both quantities are close to 0, but that doesn't mean we can deduce that $x^2 / x \to 1$ as x->0 - we know that the limit is in fact 0.

What we need here is actually that (sin h)/h -> 1 as h->0 and this is in fact the case, provided that h is measured in radians.

For info, here's what I regard as the 'standard' A level proof that the OP needs - taken from the 'bible' Bostock & Chandler. I assumed it was included in all standard A level texts,
but perhaps I'm just terribly naive

Thank you very much. I really do appreciate your help



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