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

Derivada[x_^n_, x_Symbol] := n*x^(n - 1)
Derivada[n_*x_, x_Symbol] := n
Derivada[Log[x_], x_Symbol] := 1/x
Derivada[_?NumericQ, x_Symbol] := 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]

How can I make a derivative of a polynomial fuction like:

Derivada[x^2 + 3 x, x]

Or:

Derivada[x^3 + x^2 + 3 x, x]

Another question:

How can I set to zero if I want to derive f[x] with respect to y for example.

And about the Chain Rule? How can I set this? For exemplo: Ho to derive

Exp[3x] 

Or

Sqrt[3x+1]
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2 Answers 2

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Here are some modified definitions you might find worth understanding

Derivada[x_^n_., x_Symbol] /; FreeQ[n, x] := n*x^(n - 1)
Derivada[f_, x_Symbol] /; FreeQ[f, x] := 0
Derivada[u_Plus, x_Symbol] := Derivada[#, x] & /@ u
Derivada[u_ *v_, x_Symbol] := u Derivada[v, x] + v Derivada[u, x]

Derivada[x^3 + x^2 + 3 x, x]
(* 3 + 2 x + 3 x^2 *)

Derivada[x^2 + 3 x, y]
(* 0 *)
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  • $\begingroup$ Why there is a dot . after n_ at Derivada[x_^n_., x_Symbol] /; FreeQ[n, x] := n*x^(n - 1)? $\endgroup$
    – Mateus
    Aug 30, 2018 at 19:24
  • $\begingroup$ See the help for Optional. It matches x as well as x^n, reducing the number of definitions needed. $\endgroup$
    – mikado
    Aug 30, 2018 at 19:29
  • $\begingroup$ And about the Chain Rule? How can I set this? For exemplo: Ho to derive Exp[3x] or Sqrt[3x+1] $\endgroup$
    – Mateus
    Aug 30, 2018 at 20:57
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You can approach this is the same kind of way you are trying above:

Derivada[z__ + y__, x_] := Derivada[z, x] + Derivada[y, x];
Derivada[x^2 + 3 x, x]
3 + 2 x

Note the double underscore, which indicates that the pattern can match a sequence, rather than a single element.

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  • $\begingroup$ Plus is Flat, so you could just use single underscores instead. $\endgroup$
    – Carl Woll
    Aug 30, 2018 at 20:14

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