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Math Help - [SOLVED] Derivative of function involving absolute values

  1. #1
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    [SOLVED] Derivative of function involving absolute values

    Hi, how would you find the derivative of 2x^2|x-1|
    This is calc ab but we haven't learned the derivatives of absolute values
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  2. #2
    Member OnMyWayToBeAMathProffesor's Avatar
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    I think the formula is \sqrt{u^2} with u being (x-1) but I am not sure how to apply that. Maybe someone else could throw in their input?
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  3. #3
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    let u be a function of x ...

    y = |u|

    y = \sqrt{u^2} = (u^2)^{\frac{1}{2}}

    \frac{dy}{dx} = \frac{1}{2} (u^2)^{-\frac{1}{2}} \cdot 2u \cdot \frac{du}{dx}

    \frac{dy}{dx} = \frac{u}{\sqrt{u^2}} \cdot \frac{du}{dx}

    \frac{dy}{dx} = \frac{u}{|u|} \cdot \frac{du}{dx}
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    so if the original equation was 2x^2(|x-1|)

    the derivative would be 2x^2 \cdot \frac{x-1}{|x-1|} \cdot 1
    =

    \frac{2x^3-2x^2}{|x-1|}

    correct?
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  5. #5
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    no ... you have to use the product rule
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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    I don't know why i keep making these simple mistakes. so if we want the derivative it would be 4x \cdot \frac{x-1}{|x-1|}+2x^2 ?

    I am getting stuck here.
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  7. #7
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    y = 2x^2 |x-1|

    \frac{dy}{dx} = 2x^2 \cdot \frac{x-1}{|x-1|} + 4x|x-1|
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  8. #8
    Member OnMyWayToBeAMathProffesor's Avatar
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    Thank you very much!
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