1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Total derivative of a partial derivative

  1. Nov 5, 2014 #1
    Im doing a question on functionals and I have to use the Euler lagrange equation for a single function with a second derivative. My problem is I dont know how to evaluate [itex]\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/itex]. Here y is a function of x, so [itex]y'=\frac{dy}{dx}[/itex].
    I know this is probably something I should be able to do by this stage but such is life.
     
  2. jcsd
  3. Nov 5, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    (a) To calculate [itex]\frac{\partial F}{\partial y''}[/itex], treat y'' as a variable independent of x, y, and y':[tex]
    \frac{\partial y}{\partial y''} = \frac{\partial y'}{\partial y''} = \frac{\partial x}{\partial y''}= 0 \\
    \frac{\partial y''}{\partial y''} = 1.[/tex]
    (b) To calculate the second derivative of that with respect to x, give y, y' and y'' their normal meanings: [tex]
    \frac{dy''}{dx} = y''' \\ \frac{dy'}{dx} = y'' \\ \frac{dy}{dx} = y' \\ \frac{dx}{dx} = 1.[/tex]
     
  4. Nov 5, 2014 #3
    Im a little confused by your answer. I know that [tex]\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial^2 F}{\partial y' \partial y}\frac{dy}{dx}+\frac{\partial^2F}{\partial y'^2}\frac{d^2y}{dx^2}+\frac{\partial^2 F}{\partial y'\partial x}[/tex]
    But I don't know how to extend this to find [tex]
    \frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/tex]
     
  5. Nov 5, 2014 #4

    pasmith

    User Avatar
    Homework Helper

    It's the same priniciple: to find [itex]\frac{d}{dx}(\frac{\partial F}{\partial y'})[/itex] you first calculate [itex]\frac{\partial F}{\partial y'}[/itex] by regarding y' as a variable independent of [itex]x[/itex] and [itex]y[/itex]. Then you differentiate the result with respect to [itex]x[/itex], regarding y' as the derivative of y with respect to x.

    Example: let [itex]F(x,y,y') = x + yy' + y'^2[/itex]. Then [tex]
    \frac{\partial F}{\partial y'} = y + 2y' \\
    \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = \frac{d}{dx}(y + 2y') = y' + 2y''.
    [/tex]

    It's the same principle with [itex]\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/itex]. Let [itex]F(x,y,y',y'') = x + y + y'y'' + e^{y''}[/itex]. Then [tex]
    \frac{\partial F}{\partial y''} = y' + e^{y''} \\
    \frac{d}{dx} \left(\frac{\partial F}{\partial y''} \right) = y'' + y'''e^{y''} \\
    \frac{d^2}{dx^2} \left(\frac{\partial F}{\partial y''} \right) = y''' + y^{(4)}e^{y''} + (y''')^2 e^{y''}.[/tex]
     
    Last edited: Nov 5, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Total derivative of a partial derivative
  1. Total Derivative (Replies: 2)

  2. Total derivative? (Replies: 6)

  3. Total derivative (Replies: 1)

Loading...