# How can we calculate $(\log_{x}{x})'$?

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

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

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

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

Notice that $\log_x x=1.$ Is that enough?
• 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 $b^c=a.$ – Maxim Gilula Oct 22 '13 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, $\log_x(1+x)$.