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Math Help - Differentiating y

  1. #1
    Senior Member Paze's Avatar
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    Differentiating y

    I can't seem to wrap my head around differentiating y or f(x). For example: x^2+xy^2=2.

    I understand the product rule and everyone keeps saying that ''it's just the product rule'' when I'm differentiating implicitly like that, but I don't have the intuition, it's just missing. Can someone attempt to clarify please? You do not need to use my example.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Differentiating y

    Can you show how you would use implicit differentiation on your example equation? It would help us to better help you if we see where exactly you are stuck.
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  3. #3
    Junior Member Bradyns's Avatar
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    Re: Differentiating y

    We start with the original equation:
     x^2+xy^2=2

    Which goes to,
    \frac{\mathrm{d} }{\mathrm{d} x}x^2+ \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = \frac{\mathrm{d} }{\mathrm{d} x}2

    Now, you diffentiate any 'x' variable as per normal, but when you are differentiating with respect to 'y', you leave the differential operator there, I will do so in the ' y'' form

    \frac{\mathrm{d} }{\mathrm{d} x}x^2+ \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = \frac{\mathrm{d} }{\mathrm{d} x}2

    I will do this in 3 parts.
    The first term x^2
    \frac{\mathrm{d} }{\mathrm{d} x}x^2 = 2x

    The second term xy^2
    \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = x\cdot2y\cdot y' + 1\cdot y^2

    The third term 2
    \frac{\mathrm{d} }{\mathrm{d} x}2 = 0

    All up, we get:
    2x + 2xyy' + y^2 = 0

    2xyy'  = -(y^2 + 2x)

    y' = -\frac{y^2 + 2x}{2xy}
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  4. #4
    Senior Member Paze's Avatar
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    Re: Differentiating y

    Thanks, I realize how to actually perform the calculation but I'm not interested in plugging and playing so to speak. I'm trying to understand exactly why we 'leave the differential operator there'. It has something to do with y being a function and the product rule but my brain is not making the complete connection.
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  5. #5
    Senior Member Paze's Avatar
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    Re: Differentiating y

    Basically what I'm trying to understand is this sentence: The derivative of y is y. The derivative of x is taken in the normal way of a variable, so that the derivative of x is 1, as an example. The reason for this is that y is a function of x, and does not exist without being dependent on the equation on the other side, on the other hand, x exists all along the domain. I mentioned this briefly in my overview of derivatives.
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  6. #6
    Junior Member Bradyns's Avatar
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    Re: Differentiating y

    Take this equation for example.
    x^2 + yx = 2

    It isn't explicitly a function of y, with respect to x, you have to do some manipulation before differentiating normally..

    y = ?

    In this case,

    y=\frac{2 - x^2}{x}

    If I gave you this originally, you would be able to differentiate this, and the problem is no different.

    But, one thing that we do usually forget is that we differentiate both sides.

    \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}\frac{2 - x^2}{x}

    \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}2x^{-1}-\frac{\mathrm{d}y}{\mathrm{d} x}x

    So we get,
    1\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1

    Take note of the left hand side.

    Finally we would get:
    \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1
    Last edited by Bradyns; May 4th 2013 at 05:09 AM.
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  7. #7
    Senior Member Paze's Avatar
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    Re: Differentiating y

    Quote Originally Posted by Bradyns View Post
    Take this equation for example.
    x^2 + yx = 2

    It isn't explicitly a function of y, with respect to x, you have to do some manipulation before differentiating normally..

    y = ?

    In this case,

    y=\frac{2 - x^2}{x}

    If I gave you this originally, you would be able to differentiate this, and the problem is no different.

    But, one thing that we do usually forget is that we differentiate both sides.

    \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}\frac{2 - x^2}{x}

    \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}2x^{-1}-\frac{\mathrm{d}y}{\mathrm{d} x}x

    So we get,
    1\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1

    Take note of the left hand side.

    Finally we would get:
    \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1
    so if the equation was instead:

    y^2=\frac{2 - x^2}{x}

    Would the answer be

    2\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1 ?
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  8. #8
    Junior Member Bradyns's Avatar
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    Re: Differentiating y

    No, the answer would be,

    2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1


    If it were,
    y^{3} = ????

    The derivative would be,
    3y^{2}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1

    If it were,
    y^{4} = ????

    The derivative would be,
    4y^{3}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1

    If it were,
    e^{y} = ????

    The derivative would be,
    e^{y}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1

    etc...
    Last edited by Bradyns; May 4th 2013 at 05:21 AM.
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