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The Chain Rule

  1. Jul 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of [tex]\ y=\sqrt{x^{2}-x}(x-1)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I'm not sure how to begin for this one. All I know to do is to change [tex]\sqrt{x^{2}-x}[/tex] to (x^2-x)^1/2 Do you I have to take the derivative of each one separately?
     
  2. jcsd
  3. Jul 22, 2009 #2

    rock.freak667

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    Firstly, you will see that y is of the form y=uv so you will need your product law here.

    so u=(x2-x)1/2 right?

    Now you want to get du/dx

    so the chain rule must be used now. if you put t=x2-x, then what is dt/dx equal to? Then what is u equal to in terms of t?
     
    Last edited: Jul 22, 2009
  4. Jul 22, 2009 #3
    Using the product rule I got, 1/2(x^2-x)[tex]^{-1/2}[/tex](x-1)+(x[tex]^{2}[/tex]-x)[tex]^{1/2}[/tex]. that doesn't seem right, did I mess somthing up?
     
  5. Jul 22, 2009 #4

    rock.freak667

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    I made a typo that probably confused you.

    So I will start over.

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

    so u= (x2-x)1/2
    and v= x-1 (dv/dx is easily found right?)

    The problem lies with du/dx....so we let t=x2-x and so u = ?? (in terms of t)

    to apply the chain rule now

    [tex]\frac{du}{dx}= \frac{d?}{dx} \times \frac{d??}{dt}[/tex]

    can you think of what variable '?' is and '??' is? (hint: 'd?' will cancel out with dt)
     
  6. Jul 22, 2009 #5
    dv/dx=1 but what is t?
     
  7. Jul 22, 2009 #6

    rock.freak667

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    yes

    t is a new variable we introduced for u=(x2-x)1/2

    we said t =x2-x , so u=t1/2

    from t=x2-x, what derivative can we find?

    From u=t1/2 what derivative can we find?
     
  8. Jul 22, 2009 #7
    [tex]\ \frac{dt}{dx}=2x-1[/tex]

    [tex]\ \frac{du}{dt}=1/2t^{-1/2}[/tex]

    So by multiply them together I will find dy/dx?
     
  9. Jul 22, 2009 #8

    rock.freak667

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    you will find du/dx.


    You used the product rule correctly in post#3, but du/dx was wrong.
     
  10. Jul 22, 2009 #9

    Mentallic

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    I remember when I began learning to take derivatives using the product rule, it never occurred to me to use other variables such as t when finding dy/dx, and I did just fine without it too. Only once I was very familiar with it all did I start using other variables to apply the function of a function rule.

    Stratosphere, maybe this is the best approach for you to take as well?

    Since [tex]y=\sqrt{x^2-x}(x-1)[/tex] requires the product rule:

    [tex]y=uv[/tex] then [tex]y'=u'v+v'u[/tex] where [tex]u=\sqrt{x^2-x}[/tex] and [tex]v=x-1[/tex]

    Just take each variable in the product rule separately:

    first, what is [tex]u'[/tex]? [tex]u'=\frac{d}{dx}(\sqrt{x^2-x})[/tex]
    next, what is [tex]v'[/tex]?...
    etc.
    Once you have them all, just substitute into the product rule formula.

    Remember that if [tex]y=[f(x)]^n[/tex] then [tex]\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)[/tex]
    When you tried applying the product rule earlier, you neglected the last [tex]f'(x)[/tex] bit.
     
  11. Jul 22, 2009 #10

    djeitnstine

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    Hmm I propose this method [tex] \sqrt{x(x-1)}(x-1)[/tex]

    [tex] \sqrt{x}(x-1)^{\frac{1}{2}}(x-1)^1[/tex] by the power laws we get

    [tex] x^{\frac{1}{2}}(x-1)^{\frac{3}{2}}[/tex] I believe this seems less messy to deal with.
     
  12. Jul 22, 2009 #11

    Mentallic

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    You made a typo. [tex]\sqrt{x^2-x}=\sqrt{x(x-1)}\neq\sqrt{\frac{1}{x}(x-1)}[/tex]

    and yes I agree with you, it does make the question simpler.
     
  13. Jul 22, 2009 #12

    djeitnstine

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    Yes I fixed it thanks
     
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