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Properties of Differential Operator and Proper Formalism

  1. May 19, 2012 #1
    Let us take a function defined by

    [tex]y=cx[/tex]
    To differentiate that, we use the operator d/dx
    [tex]\frac{d}{dx} y = \frac{d}{dx} cx^{-1}[/tex]
    By the chain rule/implicit differentiation on the left and normal differentiation on the right we get,
    [tex]\frac{dy}{dx} = -1c x^{-2}[/tex]

    What confuses me is the proper usage of the formalism to get

    [tex]\frac{d^2y}{dx^2} = 2cx^{-3}[/tex]

    It doesn't seem that we can use the same operation as last time.
     
  2. jcsd
  3. May 19, 2012 #2
    Your initial calculation is wrong, x is not supposed to be x^(-1), it remains x, and after the differentiation it is 1. The second derivative is then, of course, 0.
     
  4. May 19, 2012 #3

    AlephZero

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    Why not?

    ##\dfrac{d^2y}{dx^2}## is just a (slightly confusing, if you think too hard about what ##dx^2## is supposed to mean) abbreviation for ##\dfrac{d}{dx}\left(\dfrac{d}{dx}y\right)##.

    (I assumed your first equation was a typo for ## y = cx^{-1}##).
     
  5. May 19, 2012 #4
    Typo.
    Oh thank you! Now I realize why it is written in that form
     
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