Say I have

fun = D[f[x,y],{y,2}]
F = Subscript[f, 1][x,y] + \[Delta]^2 Subscript[f, 2][x,y]

I want to replace f with F so I do

fun /. f -> F

which gives

((Subscript[f, 0][x,y]+\[Delta]^2 Subscript[f, 2][x,y]+\[Delta]^4 Subscript[f, 4][x,y])^(0,2))[x,y]

Why doesn't the expansion derivative get evaluated? E.g. why is the output $\partial_y^2(f_0 + \delta^2f_2+\delta^4f_4) $ instead of $\partial_y^2f_0 + \delta^2\partial_y^2f_2 + \delta^4\partial_y^2f_4$?

Unless I missed something there should be d^2 next to the first term, or?

Here is what you can do:

fun /. f -> Function[{x, y}, d^2 g[ d x, d y]] 

or, less verbose

fun /. f -> ( d^2 g[ d #, d #2]& )

Why did your examples fail?

Nothing prevents D[f[x,y],x] from evaluation so it evaluates to Derivative[1, 0][f][x, y].

• This is why your first example fails, it does not much it.
• The second example does not work because you are replacing f, which is an operator with a product of \[Delta]^2 g where g was supposed to be an operator but how could MMA know it.

• Same story with the third example, additionaly, if you have a list of list of replacements you will get a list. You should provide a flat list of rules to get one result only.

• If you would've applied a flat list in the forth example, you'd get a correct result. But writing rules for each derivative is not something one would like to do. My solution avoids that because if you replace f in Derivative[...][f] with a function, Derivative will take care about applying the chain rule.

On a related note, how would you substitute for a derivative? Say I have

fun = D[f[x, y, t], x] + D[f[x, y, t], {x, 2}] + D[f[x, y, t], {x, 3}]

and want to make the substitution $\partial_xf \rightarrow g$ so that fun goes from $$\partial_xf +\partial_{xx}f + \partial_{xxx}f$$ to $$g + \partial_xg + \partial_{xx}g. $$

This doesn't work:

fun /. D[f[x, y, t], x] -> (g[#, #2, #3] &)

and

fun /. f^(1,0,0)->(g[#,#2,#3]&)

only replaces the one derivative.

Thanks!