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Math Help - Derivative of a Function at a specific point

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Derivative of a Function at a specific point

    If 4x^2-4xy+y^2=1 then \frac{dy}{dx} at (1,1) =

    1. (A) 2
    2. (B) \frac{1}{2}
    3. (C) 0
    4. (D) \frac{-1}{2}
    5. (E) -2


    Here's what I did:

    <br /> <br />
\frac{dy}{dx} [4x^2-4xy+y^2] = \frac{dy}{dx} [1]<br />

    <br />
8x - 4((x)(y')+(y)(1)) + 2yy' = 1<br />

    <br />
8 -4(y' + 1) + 2y' = 1<br />

    <br />
y' = \frac {-11}{-2}<br />
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  2. #2
    MHF Contributor
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    Quote Originally Posted by StarlitxSunshine View Post
    If 4x^2-4xy+y^2=1 then \frac{dy}{dx} at (1,1) =

    1. (A) 2
    2. (B) \frac{1}{2}
    3. (C) 0
    4. (D) \frac{-1}{2}
    5. (E) -2


    Here's what I did:

    <br /> <br />
\frac{dy}{dx} [4x^2-4xy+y^2] = \frac{dy}{dx} [1]<br />

    <br />
8x - 4((x)(y')+(y)(1)) + 2yy' = 1<br />

    8 -4(y' + 1) + 2y' = 1

    ...
    \textcolor{red}{8 - 4y' - 4 + 2y' = 1}

    recalculate y'

    ...
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  3. #3
    Junior Member StarlitxSunshine's Avatar
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    Ohh... Okay, so then,

    I get y' = -3/2 ?

    But that's not right...
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  4. #4
    MHF Contributor
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    Quote Originally Posted by StarlitxSunshine View Post
    Ohh... Okay, so then,

    I get y' = -3/2 ?

    But that's not right...
    your mistake (and mine) ...

    8 - 4y' - 4 + 2y' = \textcolor{red}{0}
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  5. #5
    Junior Member StarlitxSunshine's Avatar
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    Oh! >_< Wow, I can't believe I did that. x3

    Thank you :33

    Now, I get -4/-2, which is 2, which is (A). =)
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