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Math Help - another differentiation question

  1. #1
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    another differentiation question



    if we have the above, then how would the answer change if we changed
    d(v/x)/dx to:
    1) d(2v/x)/dx
    2) d(v^2/x)/dx
    3) d((v+3)/x)/dx

    thanks
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  2. #2
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    few question before we start.

    Are you able to derive the original result ? does the quotient rule mean anything to you?

    Bobak
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Richyie View Post


    if we have the above, then how would the answer change if we changed
    d(v/x)/dx to:
    1) d(2v/x)/dx
    2) d(v^2/x)/dx
    3) d((v+3)/x)/dx

    thanks
    Just use this \frac{D[\frac{u}{v}]}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}
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  4. #4
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    then would this be right:

    d(v^2/x)/dx = ( 2vx(dv/dx) - v^2 ) / x^2

    thnx
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Richyie View Post
    then would this be right:

    d(v^2/x)/dx = ( 2vx(dv/dx) - v^2 ) / x^2

    thnx
    \frac{D[\frac{v^2}{x}]}{dx}=\frac{x\cdot{2}\cdot{v}\cdot\frac{dv}{dx}-v^2}{x^2}
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by Mathstud28 View Post
    \frac{D[\frac{v^2}{x}]}{dx}=\frac{x\cdot{2}\cdot{v}\cdot\frac{dv}{dx}-v^2}{x^2}
    Just as a disclaimer...this answer is valid if v(x) is what v implied and not a constant
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