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Find Derivatives of a Function

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    If f(x)=[itex]\sqrt{x}[/itex]g(x), where g(4)=8 and g'(4)=7, find f'(4).


    2. Relevant equations
    the product rule/quotient rule


    3. The attempt at a solution
    I am having trouble getting this problem started.
    Would I solve for the x of f(x)?
     
  2. jcsd
  3. Oct 20, 2011 #2

    gb7nash

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    Forget the numbers for a second. Let's look for f'(x). What does the product rule say?
     
  4. Oct 20, 2011 #3
    The product rule is
    d/dx [f(x)*g(x)]=f(x)g'(x)+g(x)f'(x)
     
  5. Oct 20, 2011 #4

    gb7nash

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    Ok, good. Let's apply this to the problem now.
     
  6. Oct 20, 2011 #5
    f(x)=x[itex]^{\frac{1}{2}}[/itex]g(x)
    f'(x)=[itex]\frac{1}{2}[/itex]x[itex]^{-\frac{1}{2}}[/itex]g(x)
     
  7. Oct 20, 2011 #6

    gb7nash

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    Just so you don't get confused with using the same letters, we'll say the product rule is:

    d/dx[p(x)q(x)] = p(x)q'(x) + q(x)p'(x)

    This is only one part of it. Let p(x) = x1/2 and q(x) = g(x). What are you missing?
     
  8. Oct 20, 2011 #7
    f'(x)=x[itex]^{\frac{1}{2}}[/itex]7+8([itex]\frac{1}{2}[/itex]x[itex]^{-\frac{1}{2}}[/itex])
     
  9. Oct 20, 2011 #8

    gb7nash

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    You're partially skipping a step. The 7 and the 8 are what you get when you want to know what f'(4) is. We're just looking for the general derivative with respect to x. What's f'(x) now?
     
  10. Oct 20, 2011 #9
    f'(x)=X^(1/2)g'g(x)+g(x)(1/2)x^(-1/2) ?
     
  11. Oct 20, 2011 #10

    gb7nash

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    Fixed a typo you made. That looks fine. So what's f'(4)?
     
  12. Oct 20, 2011 #11
    f'(x)=X^(1/2)g'(x)+g(x)(1/2)x^(-1/2) ?
     
  13. Oct 20, 2011 #12
    16!

    Thank you very much :)
     
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