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proof of the differential of sin x Watch

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    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 x

    let h be a small increment to x

    \dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

    \dfrac{dy}{dx}=\dfrac{(\sin x\cos h + \sin h \cos x)-\sin x}{h}

    as h tends to zero

    \displaystyle\lim_{h\rightarrow 0}\;\sin x\cos h=\sin x

    \displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x=0

    so now I have 0/0 ?

    :confused:
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    Yeah, that almost always happens when you're trying to calculate derivatives by limits. You'll need to prove that \frac{\sin h}{h} \to 1.
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    (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 \frac{\sin h}{h} \to 1.
    You use the squeeze theorem, right?
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    And it is usually shown by a geometric argument that (sinh/h) tends to 1 - Google it, it's well documented.
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    (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.
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      You also need to prove that \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:
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      (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 \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?)
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      (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:
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      (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:

      \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.
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      Also, it is bad to write

      \dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

      This isn't true.

      \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 \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.
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      (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:

      \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

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

      and

      \displaystyle\lim_{h\rightarrow 0} \dfrac{\sin h \cos x}{h}=\cos x

      since

      \displaystyle\lim_{h\rightarrow 0} \dfrac{\sin h}{h}=1

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

      \dfrac{dy}{dx}=\dfrac{\sin(x+h)-\sin x}{h}

      This isn't true.

      \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.
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      (Original post by Plato's Trousers)
      latex]\displaystyle\lim_{h\rightarrow 0} \dfrac{\sin x(\cos h - 1)}{h} = 0 [/LATEX]
      Yes, but why?
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      (Original post by SimonM)
      Yes, but why?
      because as h -> 0, cos h -> 1, so the bracket disappears.
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      (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.
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      (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)
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      (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.


      (Original post by Saichu)
      You also need to prove that \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 )
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      ok, so here's my attempt. Is this correct?

      given

      \dfrac{\cos h -1}{h}

      multiply top and bottom by \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

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

      but

      \displaystyle\lim_{h\rightarrow 0}\dfrac{-\sin^2 h}{h}=-\sin h

      and

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

      so

      \displaystyle\lim_{h\rightarrow 0} \dfrac{\cos h -1}{h}=-\dfrac{\sin h}{2}

      and since

      \displaystyle\lim_{h\rightarrow 0} -\dfrac{\sin h}{2}=0

      then

      \displaystyle\lim_{h\rightarrow 0} \dfrac{\cos h -1}{h}=0
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      (Original post by Plato's Trousers)
      \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
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      (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.
     
     
     
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