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Derivative of this function?

  1. Oct 12, 2011 #1
    1. Let the function f(x) have the property that f′(x)=x+1/x−3. If g(x)=f(x^2) find g′(x).

    I've tried some steps already, however my answer is still wrong..

    g'(x)=?
    g'(x)=f'(x) at x^2 so, f'(x^2)?

    (x+1)'(x-3)-(x+1)(x-3)'/(x-3)^2

    in the end i get -4/(x-3)^2 and then I plug in x^2..
    >-4/(x^4-6x^2+9)??

    this seems to be wrong so could someone point out where my concept is flawed?
    Thanks
     
  2. jcsd
  3. Oct 12, 2011 #2

    HallsofIvy

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    I have no idea why you are finding the second derivative of f. That is not at all relevant to the question. Use the chain rule: [itex]g(x)= f(x^2)[/itex] so [itex]g'(x)= f'(x^2)(x^2)'[/itex]. You are given f', not f, so you do not need to do that derivative.
     
    Last edited: Oct 12, 2011
  4. Oct 12, 2011 #3
    Thanks so much, I don't exactly know what I was doing, either...haha I should have read the question more clearly.
     
  5. Oct 12, 2011 #4

    Ray Vickson

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    Since g(x) = f(x^2), we have g'(x) = (d/dx) f(x^2), and you can write this out using the chain rule together with your formula f'(x) = x-3 + 1/x (as you have written).

    RGV
     
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