Related to this, I am looking for a solution for:

$(\log_{x}{x})'$ = ?

...where $x$ is not 1, but positive.

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    $\begingroup$ $$\log_bb=1$$ if $b>0,\ne1$ $\endgroup$ Oct 21, 2013 at 18:54
  • $\begingroup$ @labbhattacharjee Oh, so simple and clean. Post an answer! :-) $\endgroup$ Oct 21, 2013 at 18:57
  • $\begingroup$ @labbhattacharjee Asked this answer my math teacher, and he didn't know to answer... ^_^ $\endgroup$ Oct 21, 2013 at 18:57
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    $\begingroup$ @Johnツ, if that's true then change of teacher...or of school. $\endgroup$
    – DonAntonio
    Oct 21, 2013 at 19:13
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    $\begingroup$ @DonAntonio: Mistakes happen sometimes. Have you never a dumb moment? $\endgroup$ Oct 22, 2013 at 0:56

2 Answers 2


Notice that $\log_x x=1.$ Is that enough?

  • $\begingroup$ Remember: $\log_b a$ is the number we raise $b$ to to get $a.$ In other words, $\log_b a= c$ if and only if $b^c=a.$ $\endgroup$ Oct 22, 2013 at 2:16

If you want to do it the hard way, let $f(y,z) = \log_y z = \frac{\ln z}{\ln y}$, so that your function is $g(x) = f(x,x)$. It is easy to compute that $\frac{\partial f}{\partial z}(y,z) = \frac{1}{z \ln y}.$ It is almost as easy to compute that $\frac{\partial f}{\partial z}(y,z) = \frac{1}{y} \cdot \frac{-\ln z}{\ln^2 y}$.

You have: $$ g'(x) = \frac{\partial f}{\partial y}(x,x) + \frac{\partial f}{\partial z}(x,x).$$ When you substitute the above formulas, everything cancels out and you find $g'(x) =0$.

Yes, it is somewhat silly to solve this particular problem this way. But hopefully, this will be of use to the author of the question if he wants to differentiate, say, $\log_x(1+x)$.


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