So, I have my function, such as


This function enters a big expression that contains variables x and y, as well as the function f. Now I have to differentiate that expression with respect to either x or y (or both). Evaluating the result, Mathematica will substitute each occurrence of f[x,y] with the definition above. For this particular function, its derivative will involve the factor of




When simplifying the result, how can I tell Mathematica to use this fact? I know there options to Simplify like TransformationFunctions, ExludedForms, ComplexityFunction, but the information and examples of the beginner-level on these options are rather scarce.


I think first I should have asked a simpler question.

In[1]:= Simplify[G m/r^2, G m == A]

Out[1]= A/r^2


In[2]:= Simplify[G m/r^2, G m/r^2 == A]

Out[2]= G m/r^2

Why is output not simply A?

  • 1
    $\begingroup$ very closely related: 3822 $\endgroup$
    – yohbs
    Nov 6, 2017 at 15:18

2 Answers 2


Instead of giving a DownValues for f, you could give definitions for it's derivative:

f /: Derivative[1, 0][f] = -1/(2 f[#1, #2] y)&;
f /: Derivative[0, 1][f] = x / (2 f[#1, #2] y^2)&;

Then, the following derivatives use f[x, y] in the output:

D[f[x, y], x] //TeXForm

$-\frac{1}{2 y f(x,y)}$

D[f[x, y], y] //TeXForm

$\frac{x}{2 y^2 f(x,y)}$

D[f[x, y] y + f[x, z] x + f[y, x] z, x] //TeXForm

$\frac{x z}{2 y^2 f(y,x)}-\frac{x}{2 y f(x,z)}-\frac{1}{2 f(x,y)}+f(x,z)$

  • $\begingroup$ Interesting that you get the higher derivatives for free; D[f[x, y], {x, 2}]. $\endgroup$
    – Edmund
    Nov 6, 2017 at 21:57
  • $\begingroup$ Is there a more general way? I gave a particular example, but I need a general solution, when I have f[x_, y_]:=whatever function of x and y, and I want to replace in a given expression each combination of x and y that yields this function with the symbol f. $\endgroup$
    – ThisGuy
    Nov 7, 2017 at 3:54

You can define f this way

frule = {f -> (Sqrt[1 - #1/#2] &)}

Then use it in any expression, such as

D[f[x, y], x] + f[x, y] /. frule
(*Sqrt[1-x/y]-1/(2 y Sqrt[1-x/y])*)

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