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Partial derivatives chain rule

  1. May 6, 2014 #1
    Suppose we have a function [tex] V(x,y)=x^2 + axy + y^2[/tex]
    how do we write
    [tex] \frac{dV}{dt} [/tex]


    For instance if [tex] V(x,y)=x^2 + y^2[/tex], then [tex] \frac{dV}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} [/tex]

    So, is the solution

    [tex] \frac{dV}{dt} = 2x \frac{dx}{dt} + ay\frac{dx}{dt} + ax\frac{dy}{dt} + 2y \frac{dy}{dt} [/tex]
     
  2. jcsd
  3. May 6, 2014 #2

    LCKurtz

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    Yes.
     
  4. May 6, 2014 #3

    ChrisVer

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    Just for generality, whenever you have a function [itex]F(x(t),y(t),...,z(t), t)[/itex] (it is a function of some functions of t, and also depends explicitly on t, for example:
    [itex]F= x(t)+y(t)^{2}+...+lnz(t) + (t^{3}-t^{2})[/itex]
    and you want to find its derivative, you have:
    [itex]\frac{dF}{dt}= \frac{\partial F}{\partial x}|_{y,...,z=const}\frac{dx}{dt}+\frac{\partial F}{\partial y}|_{x,...,z=const}\frac{dy}{dt}+...+\frac{\partial F}{\partial z}|_{x,y,...=const}\frac{dz}{dt} + \frac{\partial F}{\partial t}|_{x,y,...,z=const}[/itex]
     
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