Tree diagram to second derivative

[Jhenrique]

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?

$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}\frac{\partial v}{\partial t}+\frac{\partial f}{\partial y}\left(\frac{\partial y}{\partial t}+\frac{\partial y}{\partial w}\left(\frac{\partial w}{\partial q}\frac{\partial q}{\partial t} + \frac{\partial w}{\partial p}\frac{\partial p}{\partial t}\right )\right )$$

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

 

[Jhenrique]

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?

 

[Gridvvk]

Yep it is possible, this is a paper I came across some time ago when looking for the same answer.

http://www.sefi.be/conference-2012/Papers/Papers/030.pdf

Starts talking about higher order partial derivatives at bottom of page 3. It's a clever use of the product rule.

 

[Jhenrique]

Interesting! But, hard and not intuitive...

 

[Mark44]

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.

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

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.

 

[Jhenrique]

This I already know! What I don't know is w.r.t. to 2 variables. This I think hard