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Math Help - Derivatives

  1. #1
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    Derivatives

    Hi,
    can you help me with these two?

    1. Find y' of y= √[(x-1)(x+2)^2

    2. Use the quotient rule or the power rule to find y'.

    if z = 2y^2 - 4y + 5, y = 6x - 5, and x = 2t, find dz/dt when t = 1.

    Thanks folks!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by robertocusato View Post
    Hi,
    can you help me with these two?

    1. Find y' of y= √[(x-1)(x+2)^2

    2. Use the quotient rule or the power rule to find y'.

    if z = 2y^2 - 4y + 5, y = 6x - 5, and x = 2t, find dz/dt when t = 1.

    Thanks folks!

    y=((x-1)(x+2)^2)^{\frac{1}{2}}

    y' = \left(\frac{1}{2}\right)((x-1)(x+2)^2)^{\frac{-1}{2}}\left(\frac{d((x-1)(x+2)^2)}{dx}\right)

    y'=\left(\frac{1}{2}\right)((x-1)(x+2)^2)^{\frac{-1}{2}}\left((x+2)^2 + 2(x-1)(x+2)\right)

    y'=\frac{(x+2)^2 + 2(x-1)(x+2)}{(2)\sqrt{(x-1)(x+2)^2}}


    show some work try other questions and present them if there is a mistake we will fix it
    Last edited by Amer; October 2nd 2009 at 09:20 PM. Reason: I Fixed The Error
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  3. #3
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    Quote Originally Posted by Amer View Post
    y=\sqrt{(x-1)(x+2)^2}

    y=(x+2)\sqrt{x-1}=(x+2)(x-1)^{\frac{1}{2}}
    Not quite. y= |x+2|\sqrt{x-1}.

    y'= (x-1)^{\frac{1}{2}}\cdot \frac{d}{dx}(x+2) + (x+2)\cdot \frac{d}{dx}(x-1)^{\frac{1}{2}}

    y'=(x-1)^{\frac{1}{2}}(1) + (x+2)\left(\frac{1}{2}\right)(1)(x-1)^{\frac{-1}{2}}

    show some work try other questions and present them if there is a mistake we will fix it
    Last edited by mr fantastic; October 2nd 2009 at 05:36 PM. Reason: Removed a redundant quote tag
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