Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook