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Homework Help: Finding the derivative of a function

  1. Dec 23, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I got as far as

    (5-x2)(1/2)(x)^-1/2 + (x)^1/2 (-x)(-2x)

    Im using the chain rule here I believe but im a tad lost as where to go from here, if ive even done it correctly up to this point. Any help would be greatly appreciated.

  2. jcsd
  3. Dec 23, 2008 #2
    How do you get this? The derivative of x2 is 2x. So it should be [tex]\sqrt{x}[/tex](-2x).
  4. Dec 23, 2008 #3
    Im not sure, this chain rule is explained extremely poorly in my manual. Isnt the equatrion I should be using

    u * dv/dx + v * du/dx ?
  5. Dec 23, 2008 #4
    Now u dv/dx = (5-x2) ( 1/2 x^-1/2), which you've got,
    and v du/dx = [tex]\sqrt{x}[/tex] (-2x).

    Add the two parts to get dy/dx.
  6. Dec 23, 2008 #5
    I dont know where to go from there.

    I got (1/2x^-1/2)(5-x2)

    But that seems wrong.
  7. Dec 23, 2008 #6
    That's just one part. Add u dv/dx and v du/dx to get dy/dx. Then simplify the expression.
  8. Dec 23, 2008 #7
    Thats what I thought I did. I had

    1/2x^-1/2 (5 - x2 + 2x - 2x) which led me to my answer.
  9. Dec 23, 2008 #8
    How did you get this?
    I get,
    dy/dx = (5-x2)(1/(2[tex]\sqrt{x}[/tex])) - 2x3/2
    dy/dx = 5/(2[tex]\sqrt{x}[/tex]) - 1/2x3/2 -2x3/2

  10. Dec 23, 2008 #9
    I dont get how you get to that second line of dy/dx. how did 2 factors minus 1 factor = 3 factors all subtracted from each other. In fact, how you got -2x^3/2 isnt clear either.
    Last edited: Dec 23, 2008
  11. Dec 24, 2008 #10


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    The "distributive law": a(b+ c)= ab+ bc.

    [tex](5- x^2)(1/(2\sqrt{x})= 5(1/(2\sqrt{x})- x^2(1/2\sqrt{x})[/tex]

    Further, [itex]\sqrt{x}= x^{1/2}[/itex] so [itex]1/(2/sqrt{x})= (1/2)x^{-1/2}[/itex] and [itex]x^2(1/2\sqrt{x})= (1/2)x^{2- 1/2}= (1/2)x^{3/2}[/itex].
  12. Dec 24, 2008 #11
    We havent been taught that method, I imagine I wont get marks for it if I do it that way. I have to use one of the derivative rules to get the final answer.
  13. Dec 24, 2008 #12


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    Which method are you reffering to? Halls' use of the distributive property or the use of the product rule (d(uv)/dx = u'v + v'u) and power rule (d(x^n) = nx^(n-1))?
  14. Dec 24, 2008 #13
    The distributive property
  15. Dec 24, 2008 #14


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    If you've had an elementary algebra course you should be familiar with the distributive property. I don't think you would lose marks for applying an algebraic property to simplify your answers.
  16. Dec 24, 2008 #15
    From what ive gathered from the course I think I would. The professor wants an answer to the question using the rules in the unit. There must be an answer to the question using the deriviative rules isnt there?
  17. Dec 24, 2008 #16


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    Yes, but it's not in it's simplest form. I would clarify with your professor that he doesn't want you to apply algebraic rules to simplify expressions.
  18. Dec 24, 2008 #17


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    It was not a question of "derivative rules", it is a question of using algebra. I can't believe that you are not "allowed" to use what you have learned before this course. You are allowed to use arithmetic aren't you?
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