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Math Help - Need help on Implicit Differentiation.

  1. #1
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    Need help on Implicit Differentiation.

    The problem goes as such:

    4xy-2x+4=0

    Evaluate y' at the point (-2,1)

    What I did was (I use 'D' since I'm not sure on the proper notation at the moment, but I want to say it's d / dx)

    D(4xy)-D(2x)+D(4)=D(0)

    4y'-2+0=0

    \frac{4y'}{4}-2+2+0=\frac{0+2}{4}

    y'=  \frac{2}{4}
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  2. #2
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    Quote Originally Posted by Zanderist View Post
    The problem goes as such:

    4xy-2x+4=0

    Evaluate y' at the point (-2,1)

    What I did was (I use 'D' since I'm not sure on the proper notation at the moment, but I want to say it's d / dx)

    D(4xy)-D(2x)+D(4)=D(0)

    4y'-2+0=0

    \frac{4y'}{4}-2+2+0=\frac{0+2}{4}

    y'=  \frac{2}{4}

    you have to use the product rule for the first term of the equation ...

    \frac{d}{dx}(4xy) = 4xy' + 4y
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  3. #3
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    What in this case is what is:
    g(x)
    g'(x)
    f(x)
    f'(x)

    I assume f(x) is the first term.

    4xy-2x+4=0

    but I'm kinda lost after that.
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  4. #4
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    Quote Originally Posted by Zanderist View Post
    What in this case is what is:
    g(x)
    g'(x)
    f(x)
    f'(x)

    I assume f(x) is the first term.

    4xy-2x+4=0

    but I'm kinda lost after that.

    \frac{d}{dx}(4xy - 2x + 4 = 0)

    4x \cdot y' + 4y - 2 = 0

    4x \cdot y' = 2 - 4y

    2x \cdot y' = 1 - 2y

    y' = \frac{1-2y}{2x}
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    How is 4y created?
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  6. #6
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    Quote Originally Posted by Zanderist View Post
    How is 4y created?
    I told you in my first reply ... the product rule.

    (4x) is the first function, (y) is the second.

    \frac{d}{dx} (4x)(y) = (4x)(y') + (4)(y)
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  7. #7
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    Okay... just a few more things:

    \frac{d}{dx}

    Means in respect to 'x' right?

    And Implicit Differentiation is used to find a Z-axis? What is it's applications?
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  8. #8
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    Quote Originally Posted by Zanderist View Post
    Okay... just a few more things:

    \frac{d}{dx}

    Means in respect to 'x' right? yes

    And Implicit Differentiation is used to find a Z-axis? no, its used to find a derivative.

    What is it's applications? besides more advanced problems, your first application of implicit differentiation will most likely be solving related rates problems.
    ...
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  9. #9
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    Quote Originally Posted by skeeter View Post

    4x \cdot y' = 2 - 4y

    2x \cdot y' = 1 - 2y
    How does that top step become the bottom step?

    4x becoming 2x, -4y becoming -2y, and finally 2 becoming a 1.
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  10. #10
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    Quote Originally Posted by Zanderist View Post
    How does that top step become the bottom step?

    4x becoming 2x, -4y becoming -2y, and finally 2 becoming a 1.
    <br />
4x \cdot y' = 2 - 4y<br />

    <br />
4x \cdot y' = 2(1 - 2y)

    Now divide both sides of the equation by 2
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  11. #11
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    Skeeter, I think I got it now! (for this problem type)

    I was looking at what was going on in this post.

    http://www.mathhelpforum.com/math-he...161-post5.html

    So what I did was this.

    4xy-2x+4=0

    Divide everything by 2:
    2xy-x+2=0
    (I did this step because I thought it would be easier to get my end result )

    Differentiate:  \frac {dy}{dx}(2xy-x+2)=0
    (To respect of 'x' of course)


    Becoming:
    2x \frac {dy}{dx} +2y - 1 = 0

    Next: 2x \frac {dy}{dx} = 1-2y

    Final: \frac  {dy}{dx}=\frac {1-2y}{2x}


    Just tell me, did I do anything illegal.
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