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

$\dfrac{d}{dr}\ln(r/a)$

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.

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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
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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

2 Answers 2

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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

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

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

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