How can we write a function that if we input an expression f, it returns the log derivative $\frac{1}{f} \frac{df}{dx}$? We have to use a conditional or pattern test so that the function accepts any symbols as input except for lists.

I will show all my attempts and I have some questions about why certain ways are wrong and why are they wrong. I am new to Mathematica and still not very familiar about how it works.

Here are all my attempts:
In: $f = x^{2};$
In: func[f_] = $\frac{1}{f} f'$
Out: $\frac{(x^{2})'}{x^{2}}$
This attempt is wrong because first we are not using conditional or pattern test. Also, why cannot we differentiate the function by using the ' sign?

In: $f = x^{2};$
In: func[f_] := $\frac{1}{f} f'$
Out: No output
Why is there no output when I use delayed assignment?

In: $f = x^{2};$
In: func[f_,x_] = $\frac{1}{f} f'$
Out: $\frac{(x^{2})'}{x^{2}}$
Is adding another argument x_ redundant, or does it make any difference?

In: $f = x^{2};$
In: func[f_] = Module[{f}, Switch[f, _Symbol, $\frac{1}{f} f'$]] Output: $\frac{f$8344'}{f$8344}$
The syntax might not make any sense, but I am struggling to find ways to fix it. And I might have used _Symbol incorrectly, what is _Symbol after all?

Now I am trying to write the function in a less general way.
Now In: $f = x^{2};$
In: func[f_] = $\frac{1}{f} D[f,x]$
Output: $\frac{2}{x}$
This method of course works but if I change the input to f = $y^{2}$, the function will return 0 as output.
One way to solve this is by changing to func[f_] = $\frac{1}{f} D[f,y]$, but I think this is considered as cheating, since we are not really answering the question and meeting the requirements set by the question.

What really makes me struggle is to compute the derivative, how can we compute the derivative of the input function with respect to its independent variable.

Also, I don't fully understand the Switch condition:
It is mentioned in Mathematica Help, that Switch evaluates $expr_i$, then compares it with each of the $form_i$ in turn, evaluating and returning the $value_i$. What is form and what is value?
Suppose I write the function:
In: f[x_] := Switch[x, x>0, 1, x<=0, 0] In: {f[5], f[-5], f[x]}
Out: {0,0,0}
Am I using the wrong form?

Any detailed help is greatly appreciated. Thanks!

  • $\begingroup$ user, we have a more appropriate way to format Mathematica code than using LaTeX. See editing help. While editing you can select the code and press Ctrl+K to get code formatting. $\endgroup$
    – Mr.Wizard
    Commented May 16, 2013 at 5:46
  • $\begingroup$ Note that the expressions like func[f_]:=f'/f are not instructions about how to differentiate the particular function f that you've defined as "In:"! That is the definition for the pattern f_ which you just happen to have used the same letter for as your function. $\endgroup$ Commented May 16, 2013 at 14:18

2 Answers 2


Here's another way to proceed, using Derivative[], and sidestepping the use of a dummy variable:

LogDerivative[f_] := Derivative[1][Composition[Log, f]]



   PolyGamma[0, x]

LogDerivative[#^3 &][x]
  • 1
    $\begingroup$ I guess you think you deserve credit for this one? Okay, +1. ;^) $\endgroup$
    – Mr.Wizard
    Commented May 16, 2013 at 10:51
  • $\begingroup$ Well, only a bit. This won't be a blockbuster anytime soon. :) $\endgroup$ Commented May 16, 2013 at 10:57
  • 1
    $\begingroup$ If you can predict that ahead of time you understand this site better than I do. $\endgroup$
    – Mr.Wizard
    Commented May 16, 2013 at 10:58
  • $\begingroup$ I think there should be a block for accepting immediately the answers by newbies as they don't learn much if anything. $\endgroup$
    – Artes
    Commented May 16, 2013 at 11:09
  • 1
    $\begingroup$ @Artes that would only discourage them from Accepting answers at all, which we really don't want. Better (IMHO) are the gentle reminders I've often made to wait a day for best exposure. $\endgroup$
    – Mr.Wizard
    Commented May 16, 2013 at 11:55

Your operator must depend on both function and variable - in analogy to D function:

logD[f_, x_] := D[f, x]/f

or an alternative definition:

logD[f_, x_] := D[Log[f], x]

Of course your variables of differentiation and in the function must agree. Test it:

logD[f[x], x]


logD[Sin[x], x]


f = x^2; logD[f, x]


g[y_] := Exp[1/y]; logD[g[y], y]


  • $\begingroup$ I tried it and it doesn't work. I set f = $x^{2}$ and the output is $x^{2}$, which is wrong. $\endgroup$
    – user71346
    Commented May 16, 2013 at 5:09
  • $\begingroup$ @user71346 Works for me. Restart Mathematica or the kernel. You made to many assignments probably and they mess with each other. Another way would be also simply logD[x^2,x]. $\endgroup$ Commented May 16, 2013 at 5:12
  • $\begingroup$ Yes it works! Thanks! Just for the sake of curiosity, is there any other method if we only take one input f. Also can we use conditional somehow? $\endgroup$
    – user71346
    Commented May 16, 2013 at 5:21

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