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Need help finding the derivative of this function!

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    f(x)=x(x2-4)1/2



    2. Relevant equations



    3. The attempt at a solution

    so I started by, let,

    u=x

    u'=1

    v=(x2-4)1/2

    v'=1/2(x2-4)-1/2(2x)

    so using the product rule

    f'(x)= (x2-4)1/2+x(1/2(x2-4)-1/2(2x))

    =(x2-4)1/2+x2(x2-4)-1/2

    Now do I just differentiate the second term again Im not quiete understanding what needs to be done here??
     
    Last edited: Apr 28, 2012
  2. jcsd
  3. Apr 28, 2012 #2
    Where did the 4 come from?
     
  4. Apr 28, 2012 #3
    Sorry error in the question i'll edit now.
     
  5. Apr 28, 2012 #4

    Mentallic

    User Avatar
    Homework Helper

    Ok so you changed the question to [tex]f(x)=x\sqrt{x^2-4}[/tex] ?

    You have both the square root and the power to a half. Choose one or the other, but not both :tongue:

    You have a function f(x) that is comprised of the product of two other functions g(x) and h(x), so [tex]f(x)=g(x)h(x)[/tex] where g(x)=x and [itex]h(x)=\sqrt{x^2-4}[/itex]. Now, to take the derivative of f(x) we use the rule

    [tex]f'(x)=g(x)h'(x)+g'(x)h(x)[/tex]

    Or, if we let g(x)=u and h(x)=v,

    [tex]f'(x)=uv'+u'v[/tex]

    So once you've found the derivative of u and v, you plug u, v, u' and v' into the formula to obtain the derivative of f.
     
  6. Apr 28, 2012 #5
    But If you look at my working I have already done that? but I still get left with x2(x2-4)-1/2+the first term.

    but the answer says (2x2-4)/(x2-4)1/2

    for starters where does that fraction even come from?

    and when I differentiate once I don't get the answer I get

    =(x2-4)1/2+x2(x2-4)-1/2
     
  7. Apr 28, 2012 #6
    That second term should be x^2/(....) not x^2(...)

    That answer is the same, they've just rationalised. Multiply through by the denominator of the fraction, and you will get the answer they gave. It's the same thing.
     
  8. Apr 28, 2012 #7
    Ok Thanks!
     
  9. Apr 28, 2012 #8

    Mentallic

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    Homework Helper

    Yes, I know. I was trying to help you get rid of your uncertainties as to what you need to do from there :smile:
     
  10. Apr 28, 2012 #9

    OOOHHH thanks a bunch mentallic!!!:tongue2:
     
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