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Math Help - Implicit differentiation - where'd the x' go?

  1. #1
    cnb
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    Implicit differentiation - where'd the x' go?

    Book example (and everywhere else online):
    Differentiate the folium of descartes:

    x^3 + y^3 = 6xy

    Left side: 3x^2 + 3y^2 y'

    Right side using product rule: 6xy' + 6y

    Where did the x' go? (Ignoring the 6 for now):

    The product rule is (fg)' = fg' + gf'
    or:
    (xy)' = xy' + yx'


    "xy' + yx'" is not "xy' + y" -- Where the heck does the x' go?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cnb View Post
    Book example (and everywhere else online):
    Differentiate the folium of descartes:

    x^3 + y^3 = 6xy

    Left side: 3x^2 + 3y^2 y'

    Right side using product rule: 6xy' + 6y

    Where did the x' go? (Ignoring the 6 for now):

    The product rule is (fg)' = fg' + gf'
    or:
    (xy)' = xy' + yx'


    "xy' + yx'" is not "xy' + y" -- Where the heck does the x' go?
    What is the first derivative of x with respect to x?

    Or
    \left ( \frac{d}{dx} \right) x = ?

    Or if that is still confusing, let p = x. Then what is p' ?

    -Dan
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  3. #3
    cnb
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    From the book example, it would have to be 1
    Also, treating d/dx (x) as a fraction would be dx/dx = 1

    I am not all that clear on why this is though. It seems like it should be terribly obvious to me, but I am unclear on just what "the derivative of x with respect to x" really means, particularly the "with respect to" part.
    Is there a way to say "d(x) with respect to q" in Newton's prime notation?

    Thankyou for your time, even if you don't make another reply!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cnb View Post
    From the book example, it would have to be 1
    Also, treating d/dx (x) as a fraction would be dx/dx = 1

    I am not all that clear on why this is though. It seems like it should be terribly obvious to me, but I am unclear on just what "the derivative of x with respect to x" really means, particularly the "with respect to" part.
    Is there a way to say "d(x) with respect to q" in Newton's prime notation?

    Thankyou for your time, even if you don't make another reply!
    I suppose, since the prime indicates a derivative with respect to x, that the best answer to your question is that x' = 1.

    x^{\prime} = \lim_{h \to 0}\frac{(x + h) - x}{h}

    = \lim_{h \to 0}\frac{h}{h} = \lim_{h \to 0} 1 = 1

    -Dan
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  5. #5
    cnb
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    Smile

    I hadn't thought of just using the definition of derivative. That makes sense. I am sure you hear this all the time with over 6,000 posts, but I really appreciate the help!
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