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Two functions involving derivatives

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data let f be a ffunction for which f(2)=-3 and f'(x)=square root of (x^2+5) if g(x)= x^2*f(x/(x-1)), what is g'(2)?



    2. Relevant equations



    3. The attempt at a solutioni am almost positive that i need to find the opposite of derivative (antiderivative?) for f'(x), so i can find f(x) and then substitute that into g(x) to find g'(x). however. i'm not sure how to do antiderivatives. i think it is something like square root of "?" +5x, but i don't know what the antideravitive of x^2 is. i would have guessed x^3, but that would yeild 3x^2 instead of x^2.
     
  2. jcsd
  3. Oct 7, 2008 #2
    You can integrate to find f then use f(2) =-3 to find the integration constant. Then just substitute 2 in the derative of g.
     
  4. Oct 7, 2008 #3
    Umm... if you don't mid, could you try to explaint to me what the intergration method is. i don't think i've heard of it. either that or i have forgotten it. is it dealing with algebra or calculus?
     
  5. Oct 7, 2008 #4

    CompuChip

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    Or, if you don't like integrating (or don't know what it is), you can just use the chain rule to write down the derivative of g and plug in x = -2.
    Since you only do one derivation you will only get stuff with f(-2) and f'(-2), both of which you know.

    dirk's method also works but I don't see the need to find f(x) first.
     
  6. Oct 7, 2008 #5
    but in order to find g(x), don't i have to find f(x)? since f(x) is in the function of g(x)? maybe that's what you were saying at the end of your explination, but i don't really understand it.
     
  7. Oct 7, 2008 #6

    gabbagabbahey

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    Suppose you were given h(x)=2f(x) taking the derivative of that would give you h'(x)=2f'(x) would it not? Similarly, if you were given h(x)=3f(3x^2) you would get h'(x)=3f'(3x^2)*(6x). So as long as you know f', you don't need to know f in order to determine g'.
     
  8. Oct 7, 2008 #7
    ok, so i think i know what your'e saying. so in my problem, would i end up with g'(x) = 2x*f'(-1/(x-1)^2)? then plug in 2 for all x's to get g'(2)!?!?!
     
  9. Oct 7, 2008 #8

    Hootenanny

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    abbykeck: Please check your Personal Messages immediately.
     
  10. Oct 7, 2008 #9
    :redface:Uh oh... secret's out, yeah, i didn't realize i couldn't have multiple accounts. so "abbykeck" is technically still online...:blushing:
     
  11. Oct 7, 2008 #10

    CompuChip

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    Yes, something like that. I mean: the approach is all right, but I don't really think you did the derivation well. You will need both the product rule and the chain rule!
     
  12. Oct 7, 2008 #11
    wha am i using the chain rule for? the f(x/(x-1))? the product rule would be for x^2 and whatever that chain rule produces, correct? i'll try it out...
     
  13. Oct 7, 2008 #12

    CompuChip

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    g(x) is of the form a(x)*b(y(x))
    so
    g'(x) = a'(x) b(y(x)) + a(x) b'(y(x))
    where you need the chain rule for b'(y(x)) = d/dx b(y(x))
     
  14. Oct 7, 2008 #13
    g'(x)=2x(f(x/(x-1))+x^2(f(-1/(x-1)^2))?
     
  15. Oct 7, 2008 #14

    gabbagabbahey

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    Your first term is right, but you've incorrectly taken the derivative of f(x/(x-1)). Remember, the chain rule gives:

    [tex]\frac{d}{dx} f(u)=f'(u) \frac{du}{dx}[/tex]

    In your case, [itex] u=\frac{x}{x-1}[/itex]
     
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