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Math Help - Implicit differentiation. xy=1.

  1. #1
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    Implicit differentiation. xy=1.

    I'm taking a quote from my textbook:

    "We can think of the equation xy=1 as stating that two functions of x (namely, xy and 1) are equal. It follows that the derivatives of these functions are equal, so

    x\frac{dy}{dx}+y=0"


    This is stated very plainly and I am finding that I'm not following what is being said here. Maybe its the notation that I don't understand. I'm trying to figure out how xy=1 transformed into the equation set to zero. For some reason y is left as y. I'm seeing sometimes that y gets changed to \frac{dy}{dx}. I'm trying to follow my textbook the best I can. Can someone explain this to me?

    Thanks in advance...
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  2. #2
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    Re: Implicit differentiation. xy=1.

    Hey sepoto.

    The basic idea is to use the product and the chain rule. We have as follows:

    d/dx [xy] = dx/dx*y + x*dy/dx = d/dx (1) = 0 by the product rule d/dx [uv] = du/dx*v + u*dv/dx. This gives us

    y + x*dy/dx = 0.

    Remember the chain rule that d/dx u(v(x)) = dv/dx * du(v(x))/dx.
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  3. #3
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    Re: Implicit differentiation. xy=1.

    How is \frac{dx}{dx}*y change to just y?
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  4. #4
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    Re: Implicit differentiation. xy=1.

    The derivative of x with respect to x is 1, and 1\cdot y = y.
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  5. #5
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    Re: Implicit differentiation. xy=1.

    And, of course, the equation is "set equal to 0" because the derivative of the constant, 1, on the right, is 0.
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