Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Properties of Differential Operator and Proper Formalism

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

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
    Oh thank you! Now I realize why it is written in that form
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook