Derivative of unknown function with respect to global defined function

Question

I have the following code:

x[t_]:=y[t]+2*t
MyRule = D[z[x[t]],x[t]]>0 

y[t] and z[x[t]] are unknown functions.

I get the following error message: "D: 2t+y[t] is not a valid variable". I suspect the message tells me that I can not take derivitates with respect to another function. But then what should I do? I know the easy solution here will be to just take the derivative with respect to t and y using the chain rule, but my real code consist of a lot of functions that depends on other functions why I really want this specific framework to work.

Answer 1

Assume we have a function x[t] that depends an a function y[t]. And you would like to get the derivative of x[t] relative to y[t]. Toward this end, you could replace y[t] by a local variable, e.g. tmp, calculate the derivative relative to tmp. And finally replace tmp by y[t]:

myD[x_, y_] := Module[{tmp},
  D[x /.  y -> tmp, tmp] /. tmp -> y
]

With this e.g.:

x[t_] := 3 y[t]^2 + 2*t
myD[x[t], y[t]]

6 y[t]

Answer 2

The standard method of variational derivative of an axpression a wrt. to any expression b, that can be identified in a by a tree search is

D[  a/.{b -> $x},$x ]/.{$x->b}

here

x[t_]:=y[t]+2*t
MyRule = (D[ z[x[t]] /. {y[t] -> x - 2 t}, x] /. {x -> x[t]}) > 0

 Derivative[1][z][2 t + y[t]] > 0

This kind of derivative by lookup of occurencies of the expression b in the expression tree of a, just as it is presented, does not take care of any algebraic identities.

Use of such procedures with Replace makes sense only for function symbols without any additional algebraic entanglements with other symbols.

Especially for derivatives of a bunch of functions of one variable, follow the classical notational path of Newton/Leibniz/Euler of switching between symbolic function names and the same symbols with argument attached:

 vars= Union[Cases[expr, f_[t]:>f,Infinity]]

 Grad[ expr/.{f_[t]:>f}, vars]/.{ Rule@@@( Transpose[  {vars, Through[vars[t]] } ] )}