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proof of the differential of sin x

I was trying to get tot he derivative of sin x from first principles. I realise that most proofs of this use a different expansion for sin(a+b), but it should work with this one too presumably?

Could do with some help...

let
y=sin⁡xy=\sin x

let h be a small increment to x

dydx=sin⁡(x+h)−sin⁡xh\dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

dydx=(sin⁡xcos⁡h+sin⁡hcos⁡x)−sin⁡xh\dfrac{dy}{dx}=\dfrac{(\sin x\cos h + \sin h \cos x)-\sin x}{h}

as h tends to zero

lim⁡h→0  sin⁡xcos⁡h=sin⁡x\displaystyle\lim_{h\rightarrow 0}\;\sin x\cos h=\sin x

lim⁡h→0  sin⁡hcos⁡x=0\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x=0

so now I have 0/0 ?

:confused:
(edited 13 years ago)

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Reply 1
Avatar for Zhen Lin
Zhen Lin
12
Yeah, that almost always happens when you're trying to calculate derivatives by limits. You'll need to prove that sin⁡hh→1\frac{\sin h}{h} \to 1.
Original post by Zhen Lin
Yeah, that almost always happens when you're trying to calculate derivatives by limits. You'll need to prove that sin⁡hh→1\frac{\sin h}{h} \to 1.


You use the squeeze theorem, right?
Reply 3
Avatar for Swayum
Swayum
18
And it is usually shown by a geometric argument that (sinh/h) tends to 1 - Google it, it's well documented.
Reply 4
Avatar for Swayum
Swayum
18
Original post by boromir9111
You use the squeeze theorem, right?


If that's the same as the Sandwich theorem (i.e. if a < b < c and a and c tend to 0, then b tends to 0), then yeah.
Reply 5
Avatar for Saichu
Saichu
11
You also need to prove that lim⁡h→0cos⁡h−1h=0\displaystyle \lim_{h \to 0}\frac{\cos h - 1}{h } = 0. Hint: Multiplication by a conjugate.




Edit:
Original post by Plato's Trousers
x


If you're that interested in alternate proofs, a completely different proof can be found on page 21 :p:
(edited 13 years ago)
Original post by Zhen Lin
Yeah, that almost always happens when you're trying to calculate derivatives by limits. You'll need to prove that sin⁡hh→1\frac{\sin h}{h} \to 1.


yes, I have seen the "official" proofs. But I guess my question was why it doesn't work this way?

(don't all roads lead to Rome?)
Original post by Swayum
If that's the same as the Sandwich theorem (i.e. if a < b < c and a and c tend to 0, then b tends to 0), then yeah.


:yes:
Reply 8
Avatar for Swayum
Swayum
18
Original post by Plato's Trousers
yes, I have seen the "official" proofs. But I guess my question was why it doesn't work this way?

(don't all roads lead to Rome?)


It does work. But when you encounter 0/0, you hit a road block. 0/0 could be anything! You can't then just plug in h = 0 into the numerator and denominator and hope that life just works out. You have to use a different argument to show what the limit is. What we're suggesting is that you split your expression:

(sin⁡xcos⁡h+sin⁡hcos⁡x)−sin⁡xh=sin⁡x(cos⁡h−1)h+sin⁡hcos⁡xh\dfrac{(\sin x\cos h + \sin h \cos x)-\sin x}{h} = \dfrac{\sin x(\cos h - 1)}{h} + \dfrac{\sin h \cos x}{h}

And take limits separately on those two terms.
(edited 13 years ago)
Reply 9
Avatar for Swayum
Swayum
18
Also, it is bad to write

dydx=sin⁡(x+h)−sin⁡xh\dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

This isn't true.

dydx=lim⁡h→0  sin⁡(x+h)−sin⁡xh\dfrac{dy}{dx}=\lim_{h \rightarrow 0}\;\dfrac{\sin(x+h)-\sin x}{h} is true

If you don't want to write lim all the time, then you say δyδx=sin⁡(x+h)−sin⁡xh\dfrac{\delta y}{\delta x}=\dfrac{\sin(x+h)-\sin x}{h}

And take limits on both sides at the end saying that delta y/delta x tends to dy/dx.
(edited 13 years ago)
Original post by Swayum
It does work. But when you encounter 0/0, you hit a road block. 0/0 could be anything! You can't then just plug in h = 0 into the numerator and denominator and hope that life just works out. You have to use a different argument to show what the limit is. What we're suggesting is that you split your expression:

(sin⁡xcos⁡h+sin⁡hcos⁡x)−sin⁡xh=sin⁡x(cos⁡h−1)h+sin⁡hcos⁡xh\dfrac{(\sin x\cos h + \sin h \cos x)-\sin x}{h} = \dfrac{\sin x(\cos h - 1)}{h} + \dfrac{\sin h \cos x}{h}

And take limits separately on those two terms.


aha! I see.

So

lim⁡h→0sin⁡x(cos⁡h−1)h=0\displaystyle\lim_{h\rightarrow 0} \dfrac{\sin x(\cos h - 1)}{h} = 0

and

lim⁡h→0sin⁡hcos⁡xh=cos⁡x\displaystyle\lim_{h\rightarrow 0} \dfrac{\sin h \cos x}{h}=\cos x

since

lim⁡h→0sin⁡hh=1\displaystyle\lim_{h\rightarrow 0} \dfrac{\sin h}{h}=1

Thanks
Original post by Swayum
Also, it is bad to write

dydx=sin⁡(x+h)−sin⁡xh\dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

This isn't true.

dydx=lim⁡h→0  sin⁡(x+h)−sin⁡xh\dfrac{dy}{dx}=\lim_{h \rightarrow 0}\;\dfrac{\sin(x+h)-\sin x}{h} is true



Sorry. Yes, I knew that. I was being lazy.
Reply 12
Avatar for SimonM
SimonM
18
Original post by Plato's Trousers
latex]\displaystyle\lim_{h\rightarrow 0} \dfrac{\sin x(\cos h - 1)}{h} = 0



Yes, but why?
Original post by SimonM
Yes, but why?


because as h -> 0, cos h -> 1, so the bracket disappears.
Reply 14
Avatar for SimonM
SimonM
18
Original post by Plato's Trousers
because as h -> 0, cos h -> 1, so the bracket disappears.


Yeah, see that logic is wrong. I think you need to read the thread again to try to get to the bottom of how limits work.
Reply 15
Avatar for Swayum
Swayum
18
Original post by Plato's Trousers
because as h -> 0, cos h -> 1, so the bracket disappears.


But the denominator is going to 0 as well, so you're stuck in a 0/0 situation again! Your argument would be fine if we were dealing with (1 - cosh)/(1 + h): over here you can let h = 0 in the numerator and denominator and conclude the limit is 0, but not when you get 0/0 (or infinity/infinty). That's what Simon's getting at.

(sorry about being pedantic before, btw, just wanted to make sure you knew)
(edited 13 years ago)
Original post by SimonM
Yeah, see that logic is wrong. I think you need to read the thread again to try to get to the bottom of how limits work.


oh. :frown:


Original post by Saichu
You also need to prove that lim⁡h→0cos⁡h−1h=0\displaystyle \lim_{h \to 0}\frac{\cos h - 1}{h } = 0. Hint: Multiplication by a conjugate.


..ahh, you mean this? Ok, I'll try to prove that then.


Original post by Swayum

(sorry about being pedantic before, btw, just wanted to make sure you knew)


no probs. Pedantic is good. I like pedantic (otherwise I wouldn't be doing maths :smile: )
ok, so here's my attempt. Is this correct?

given

cos⁡h−1h\dfrac{\cos h -1}{h}

multiply top and bottom by cos⁡h+1\cos h +1


cos⁡2h−1h(cos⁡h+1)=−sin⁡2hh(cos⁡h+1)\dfrac{\cos^2 h -1}{h(\cos h +1)}=\dfrac{-\sin^2 h}{h(\cos h +1)}

divide top and bottom by h

−sin⁡2hhcos⁡h+1\dfrac{\frac{-\sin^2 h}{h}}{\cos h+1}

but

lim⁡h→0−sin⁡2hh=−sin⁡h\displaystyle\lim_{h\rightarrow 0}\dfrac{-\sin^2 h}{h}=-\sin h

and

lim⁡h→0cos⁡h+1=2\displaystyle\lim_{h\rightarrow 0} \cos h+1=2

so

lim⁡h→0cos⁡h−1h=−sin⁡h2\displaystyle\lim_{h\rightarrow 0} \dfrac{\cos h -1}{h}=-\dfrac{\sin h}{2}

and since

lim⁡h→0−sin⁡h2=0\displaystyle\lim_{h\rightarrow 0} -\dfrac{\sin h}{2}=0

then

lim⁡h→0cos⁡h−1h=0\displaystyle\lim_{h\rightarrow 0} \dfrac{\cos h -1}{h}=0
(edited 13 years ago)
Reply 18
Avatar for SimonM
SimonM
18
Original post by Plato's Trousers

lim⁡h→0sin⁡2hh=−sin⁡h\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin^2 h}{h}=-\sin h


This is nonsense. h is a dummy variable, it can't still appear in the RHS. If you can fix this, then yes, you're heading in the right direction
Original post by SimonM
This is nonsense. h is a dummy variable, it can't still appear in the RHS. If you can fix this, then yes, you're heading in the right direction


actually, yes, I was just re-reading it and wasn't happy with that part. :frown:

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