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Tree diagram to second derivative

  1. Nov 26, 2013 #1
    Is it possible to formulate the second derivate trough of a tree diagram, as we do with a first derivative? If yes, how do it?


    [tex]\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}\frac{\partial v}{\partial t}+\frac{\partial f}{\partial y}\left(\frac{\partial y}{\partial t}+\frac{\partial y}{\partial w}\left(\frac{\partial w}{\partial q}\frac{\partial q}{\partial t} + \frac{\partial w}{\partial p}\frac{\partial p}{\partial t}\right )\right )[/tex]

    Edit: example: suppose you need to make the second partial derivative ∂²f/∂s∂t

    Attached Files:

  2. jcsd
  3. Nov 27, 2013 #2
    I was sleepy when wrote this topic, sorry!

    My ask is: Is it possible to formulate the frist derivate trough of a tree diagram, so that, start at f and finish at t, multiplicand all derivatives contained in this path and adding all paths with this same process (I think/hope you already know how do it). So, how can we do to derive twice (with respect to different variables) using a tree driagm as tool? Is it possible?
  4. Dec 1, 2013 #3
    Yep it is possible, this is a paper I came across some time ago when looking for the same answer.


    Starts talking about higher order partial derivatives at bottom of page 3. It's a clever use of the product rule.
  5. Dec 2, 2013 #4
    Interesting! But, hard and not intuitive...
  6. Dec 2, 2013 #5


    Staff: Mentor

    Start with the first figure, where z is a function of x and y, and x and y are each functions of s and t. Then z has a partial with respect to s and a partial with respect to t.

    The partial of z w.r.t. (with respect to) s involves both paths from z to s. The things along a given path are multiplied, and the resulting products of the two paths are added.

    $$ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} $$

    The first term in the sum represents the path from z through x to s; the second term represents the path from z through y to s.

    Calculating ##\frac{\partial z}{\partial t}## would be done in a similar manner.

    Figure 2 in the linked document is doing the same thing as in Figure 1, except that it reproduces the nodes at s and t to spread the figure out.
  7. Dec 3, 2013 #6
    This I already know! What I dont know is w.r.t. to 2 variables. This I think hard
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