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Use definition of derivative

  1. Mar 15, 2005 #1
    To find the derivative of sin(2x). So far here is what I did:

    lim h -> 0 (sin(2x+2h)-sin(2x))/h
    = (sin2xcos2h+cos2xsin2h-sin2x)/h
    = (2sinxcosx(cos2h-1)+2sinhcoshcos2x)/h

    I'm stuck here... can't remember how to simplify this. Thanks. I also left out some steps, I don't know how to use the latex graphics yet and didn't want it to get to garbled.
  2. jcsd
  3. Mar 15, 2005 #2
    Are you allowed to use the fact that

    [tex] \lim_{x\rightarrow 0} \frac{\sin{\alpha x}}{\alpha x} = 1 \ \mbox{if} \ \mathbb{R} \ni \alpha \neq 0[/tex]

    ? If so, it's completely trivial. Just use [tex]\sin{x} - \sin{a} = 2\cos{\frac{x+a}{2}} \sin{\frac{x-a}{2}}[/tex].
  4. Mar 15, 2005 #3


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    If you know
    lim h -> 0, (sin(x+h)-sin(x))/h = cos(x),

    lim h -> 0,
    = (2/2)*(sin(2x+2h)-sin(2x))/h ... that is, multiply by 1 in a funny way
    = 2(sin(2x+2h)-sin(2x))/(2h)
    = 2(sin(X+H)-sin(X))/(H) where X=2x and H=2h (note that lim h->0 implies lim H->0)

    = 2cos(X)
    = 2cos(2x)
  5. Mar 15, 2005 #4
    Haha, somehow I got the idea that you wanted to find the derivative of [tex]\sin{x}[/tex] (which is actually "harder"). All you need for [tex]\sin{2x}[/tex] is

    [tex]\lim_{x\rightarrow 0} \frac{\sin{x}}{x} = 1[/tex]
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