Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Doing some rudimentary mental differentiation in my fluid mechanics homework, I encountered the following derivative:


I applied the chain rule, said it equaled $\dfrac{1}{ar}$

Needless to say, my final answer had this term off by a factor of 1/a and a bit of research showed me that the actual derivative evaluates to 1/r instead. Thinking about the properties of logarithms, this makes sense because the $\ln(a)$ can be pulled out as a constant, subtracted term that disappears in the derivative.

But my question asks for the calculus explanation of this. Why does the general form of $\dfrac{d}{dx}f(ax)=af'(ax)$ not seem to hold in this case without first expanding the logarithm? It's been years since I had single variable calculus and I have a vague recollection that something like this happens but I can't remember why for the life of me.

share|improve this question
Your application of chain rule is incorrect. Chain rule does give $1/r$ as the derivative of $\log (r/a)$. –  Srivatsan Nov 14 '11 at 15:49
If you must, a proper chain rule application should yield $\dfrac1{r/a}\cdot\dfrac1{a}$, and the cancellation should now be obvious... –  J. M. Nov 14 '11 at 15:52
@J.M. Oops! So obvious now. I really should have caught that seeing as my general form includes f'(ax) which indicates I even thought about the constant having to persist. This late in term my brain is just getting fuzzy I suppose. –  William Grobman Nov 15 '11 at 7:41
add comment

2 Answers 2

up vote 3 down vote accepted

By the chain rule:

$$\frac{d}{dr} \ln(r/a) = \frac{1/a}{r/a} = \frac{1}{r}$$

You can do the derivative another way by using properties of logarithms to first write $\ln(r/a) = \ln r - \ln a$. Then

$$\frac{d}{dr} (\ln r - \ln a) = \frac{1}{r} - 0 = \frac{1}{r}$$

share|improve this answer
add comment

The chain rule does hold. Observe: $$ \frac{d}{dr} \ln(r/a) = \frac{1}{(\frac{r}{a})} \frac{1}{a} = \frac{1}{r} $$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.