I want to create my own differentiation function Derivada. So, I already set some properties like:

Derivada[x_^(n_: 1), x_Symbol] /; FreeQ[n, x] := n*x^(n - 1)
Derivada[n_*x_, x_Symbol] /; FreeQ[n, x] := n
Derivada[Log[a_: E, x_], x_Symbol] := 1/(x*Log[a])
Derivada[Log[x_], x_Symbol] := 1/x
Derivada[f_, x_Symbol] /; FreeQ[f, x] := 0
Derivada[(a_?NumericQ) f_, x_Symbol] := a*Derivada[f, x]
Derivada[Exp[x_], x_Symbol] := Exp[x]
Derivada[a_^x_, x_Symbol] := a^x Log[a]
Derivada[u_Plus, x_Symbol] := Derivada[#, x] & /@ u
Derivada[u_*v_, x_Symbol] := u Derivada[v, x] + v Derivada[u, x]
Derivada[u_/v_, x_Symbol] := (Derivada[u, x]*v - u*Derivada[v, x])/v^2

I have a problem with rational functions.

The functions works for:

In[1224]:= Derivada[x/(x + 1), x]

Out[1224]= 1/(1 + x)^2

But the function doesn't works for:

In[1225]:= Derivada[1/(x + 1), x]

Out[1225]= Derivada[1/(1 + x), x]

What can be wrong?

  • $\begingroup$ just add the rule Derivada[1 /v_, x_Symbol] := -Derivada[v, x]/v^2? $\endgroup$
    – kglr
    Aug 30, 2018 at 23:05
  • 1
    $\begingroup$ Hi, I edited your question to improve the English a bit. Please have a look, and check that you're happy with the edit. Otherwise, you can always choose to rollback. $\endgroup$
    – QuantumDot
    Aug 31, 2018 at 15:16
  • $\begingroup$ Note that the derivative of x/(x+1) is -x/(1+x)^2 + 1/(1+x), NOT 1/(1+x)^2 as in your result! $\endgroup$
    – evanb
    Aug 31, 2018 at 15:39

1 Answer 1


The problem: Mathematica manages to change the form of the underlying expression, and the expression no longer matches the pattern.

u_/v_ // FullForm
1/(x + 1) // FullForm

It also can not match the rule Derivada[x_^(n_), x_Symbol] as in the case of Power[Plus[1,x],-1] the first argument is not the same x as the second argument in your rule (for a good reason).

The solution: You could keep on implementing a rule for every special case or you could implement a chain rule. Simplification of symbolic expressions is a field of research and sooner or later you will start publishing your methods in journals :)


Just because it is a fun excuse to take the pattern matching to the limits, here is an implementation of 2 level chain rule:


Derivada[1/(x + 1), x]

enter image description here

Next step would be to make it recursive. It should check, if any of the arguments is also a function of x, and if it is possible to take a derivative of them. It is a deep rabbit hole.

  • $\begingroup$ Some suggestion to implement the chain rule? $\endgroup$
    – Mateus
    Aug 30, 2018 at 23:07
  • 1
    $\begingroup$ Please see the updated answer. $\endgroup$
    – Johu
    Aug 31, 2018 at 0:26

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