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N order derivates

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    The exercise goes "Determine d^n*f/dx^n for the 2 exercises:"
    a) f(x)=sin^4(x) + cos^4(x)
    b) f(x)= x^n/(1-x)


    2. Relevant equations

    3. The attempt at a solution
    The only idea i had was for the second example, where i think its right to use a rule from Leibniz but I'm not sure...
     
  2. jcsd
  3. Nov 24, 2009 #2

    Mentallic

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    I've never done any of these before, so please forgive me if I'm way off and not being helpful at all.


    a) couldn't you express [itex]sin^4x[/itex] and [itex]cos^4x[/itex] in terms of multiple angles without powers, and then take the derivatives of those?


    b) you can make the fraction like this: [tex]\frac{-(1-x^n)+1}{1-x}[/tex]

    and then they split up as so: [tex]\frac{-(1-x^n)}{1-x}+(1-x)^{-1}[/tex]

    The first fraction can be... expanded (for lack of a better word) into [itex]-(1+x+x^2+...+x^{n-1})[/itex] and I'm sure it's clear what to do with the second fraction. :smile:
     
  4. Nov 24, 2009 #3
    i don't know about the first one with the angles, i think the whole point of the exercise is to keep the sin and cos and the second one kinda confused me...
     
  5. Nov 24, 2009 #4

    Mentallic

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    Well you would still keep the sin and cos, just that theyre expressed as

    [tex]sin^nx=Asin(nx)+Bsin((n-1)x)...[/tex]

    I know you could find the relationship easily for the n-th derivative from that, but if that isn't allowed, then maybe this will help?

    [tex]y=sin^4x[/tex]

    [tex]\frac{dy}{dx}=4sin^3xcosx[/tex]

    [tex]\frac{d^2y}{dx^2}=-16sin^4x+12sin^2x[/tex]

    Notice how the [itex]sin^4x[/itex] appears again. Maybe it's something, probably it's not, but I'm just putting that out there in case it helps.


    For the second, is that "confused me" or "still confuses me"?

    [tex](x^n-1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1)[/tex]

    dividing both sides by [itex]x-1[/itex] will show you how to get the long expansion thing.

    Was this the problem?
     
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