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Use the Approximate Relationship to Prove

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the approximate relationship to prove:
    Δf~[itex]\frac{df}{dx}[/itex]Δx

    a) [itex]\frac{dx^{n}}{dx}[/itex]= nxn-1

    b)[itex]\frac{dcos(θ)}{dθ}[/itex]=-sin(θ)

    2. Relevant equations
    a)N/a?? I'm not sure if I need any other equations than the one given.

    b) sin(ε)~ε and cos(ε)~1 when ε<<1.
    3. The attempt at a solution
    a) So I'm honestly quite lost and know that my attempt is going to be far off.
    I thought maybe I could sub in the values into the given equation:

    f2-f1~df/dx (x2-x1) (because they are deltas)
    xn-x~[itex]\frac{dx^{n}}{dx}[/itex](Δn)
    xn-1~[itex]\frac{dx^{n}}{dx}[/itex](Δn)

    But now I have no clue what I'm doing again and I know what I'm doing doesn't make much sense. I honestly haven't even attempted part (b) because I don't understand what to do with part (a). Just to be clear I'm not looking for help on part (b) yet until I try to attempt the problem. I'm just looking for help with part (a) so that I can try to do part (b) afterwards. Can someone help explain to me how I am suppose to use that formula?
     
  2. jcsd
  3. Sep 22, 2013 #2

    fzero

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    For part a we have

    $$\Delta f = f(x+\epsilon) - f(x) = (x+\epsilon)^n - x^n.$$

    Now you want to expand ##(x+\epsilon)^n## and then take the limit that ##\epsilon \rightarrow 0##. There's an identity from precalc that you can use here. For part b, you will have to expand ##\cos(x+\epsilon)## using trig identities.
     
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