Every time i see a logarithmic function and if it so happens that i'am required to take the derivative or the integral of that particular function i get stumped and i tend to avoid that problem.

What i'am saying is $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \log _{10}x \right )$$ and $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \ln x \right )$$ are the same i.e $$\frac{1}{x}$$ Why is that? Infact they both are different right? One is to the base 10 and the other one is to the base e ,the logarithm with base e is called as the Naperian or natural logarithm and what's the log base 10 called as?unnatural or artificial logarithm? They are required to have different derivatives right?

Is that because every property of logarithms holds good for logs of both the bases? Is there any particular reason for that?

These are some of the basics on which i need clarity before i move on with my mathematical studies.I'll be very happy if someone could give me an insight on these aspects.

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    $\begingroup$ $\log_{10} x = (\ln x)/(\ln 10)$, so how can the derivatives be equal? $\endgroup$ Sep 8, 2011 at 7:16
  • $\begingroup$ I've no clue.If you say that the question itself is incorrect then i don't know how to rephrase it. $\endgroup$
    – alok
    Sep 8, 2011 at 7:19
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    $\begingroup$ That's what I'm saying, the derivatives are in fact not equal. You must have made some mistake somewhere. Or you are confused about something. $\endgroup$ Sep 8, 2011 at 7:21

2 Answers 2


There are two mistakes in your arguments.

First, the derivatives of $\log_{10}x$ and $\ln x$ are not the same. Since $\log_{b}x=\ln x/\ln b$, we have

$$\frac{\mathrm d}{\mathrm dx}\log_{b}x=\frac{\mathrm d}{\mathrm dx}\frac{\ln x}{\ln b}=\frac1{\ln b}\frac{\mathrm d}{\mathrm dx}\ln x=\frac1{\ln b}\frac1x\;.$$

Second, it's not true that different functions must have different derivatives. Functions that differ only by an additive constant have the same derivative. To use an example close to yours,

$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=c\frac1{cx}=\frac1x\;.$$

This might seem mysterious in that form, but is becomes clearer if you instead write

$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=\frac{\mathrm d}{\mathrm dx}\left(\ln x+\ln c\right)=\frac{\mathrm d}{\mathrm dx}\ln x=\frac1x\;,$$

which shows that this is due to the fact that $\ln (cx)$ and $\ln x$ only differ by an additive constant.

  • $\begingroup$ That gave me little of understanding ,but my headache's not gone yet! I'll wait if i can get some more answers. $\endgroup$
    – alok
    Sep 8, 2011 at 7:27
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    $\begingroup$ @alok: You can wait, or you could tell us what's still giving you a headache; perhaps that might shorten the wait :-) $\endgroup$
    – joriki
    Sep 8, 2011 at 7:28

The derivatives aren't identical. Here the fact you need to note is that

$$ log_a\ b = \frac{log\ b}{log\ a}, $$

which always holds true. You can for instance take $log$ here to be the natural logarithm.

About your confusion on the specialty of natural logarithm: If you wish to see the base $e$ as natural, you might wish to define logarithm rigorously. For instance you could define natural logarithm of a positive real number $x$ as

$$log\ x= \int_1^x \frac{dt}{t} $$

or you can define it as a series, with $|x|< 1$, as:

$$log(1 +x) = \frac{x}{1} - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \cdots .$$

In both of these cases, with base '$e$' the expressions are most simple and natural, if you would like.


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