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I don't have an extensive formal training in calculus, but I'm doing quite a lot of differential calculus work at the moment and there's something which bothers me.

Say I have the differential

$\dfrac{d\ln (P(x))}{d\ln(x)}$

Where $P$ is some unknown function

By defining $u = \ln(x)$ I can re-write this as

$\dfrac{d\ln (P(e^u))}{du}$

Which I would then solve (using prime notation) as follows

$\dfrac{f(g(h(u)))}{du}$

$f(z) = \ln(z)$ , $f'(z) = \dfrac{1}{z}$

$g(z) = P(z)$, $g(z)' = P'(z)$

$h(z) = e^z$, $h'(z) = e^z$

So using the chain rule (note I'm being super explicit in my steps)

$\dfrac{f(g(h(u)))}{du} = f'(g(h(u)) \times g(h(u))' $

$g(h(u))' = g'(h(u)) \times h'(u)$

$\dfrac{f(g(h(u)))}{du} = f'(g(h(u)) \times g'(h(u)) \times h'(u) $

$\dfrac{f(g(h(u)))}{du} = \dfrac{1}{P(e^u)} \times P'(e^u) \times e^u $

$\dfrac{f(g(h(u)))}{du} = \dfrac{P'(e^u)e^u }{P(e^u)} $

$\dfrac{f(g(h(x)))}{dx} = \dfrac{xP'(x)}{P(x)}$

$\dfrac{d\ln (P(x))}{d\ln(x)} = \dfrac{x}{P(x)}P'(x)$

1) is this kind of mixing and matching of Leibniz and prime notation even "legal"?

If not, should I really re-write the differential as

$V'(\ln(x)) = \dfrac{d\ln (P(x))}{d\ln(x)} \text{ where } \Bigg(V(x) = \ln (P(x))\Bigg)$ (i.e. is this correct?)

2) I prefer to use prime notation, but does this have any significant drawbacks (other than being less "clean" in places, i.e. you can treat $\dfrac{df(x)}{dx}$ as a fraction in certain conditions).

3) given that P is a function (say P is defined as $P = kx + cx^2$), what is the difference between

$\dfrac{P(x)}{dx}$ and $\dfrac{P}{dx}$

I'm trying to work out if the stuff I'm reading is just using inconsistent notation, or if there are crucial functional/semantic differences I'm not understanding.

UPDATE: Dealing with $\dfrac{d\ln(P(x))}{d\ln(x)}$

For a concrete example, say P(x) = 2x+5, so, as mentioned above, we define $u = \ln(x)$

We would re-write $\dfrac{d\ln(2x+5)}{d\ln(x)}$ as $\dfrac{d\ln(2e^u+5)}{du}$, then solve this getting

$\dfrac{2e^u}{2e^u+5}$

And finally substitute $\ln(x)$ back in for $u$

$\dfrac{2x}{2x+5}$

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2  
I'm confused by the very first of your notation: $\frac{d\ln(P(x))}{d\ln(x)}$. Can you really write $d\ln(x)$ in the denominator? I have never come across this notation so far ... –  StringerBell Jan 21 '13 at 19:25
    
I don't see any reason why not? I mean it's maybe not a common way of writing a differential, but pops up a lot in engineering/thermodynamics (although for the record I'd never seen it before either until fairly recently). –  Alex Jan 21 '13 at 19:28
1  
What does $\frac{d\ln{P(x)}}{\ln{x}}$ mean? How is it defined? –  user39280 Jan 21 '13 at 19:31
    
Does $\frac{\mathrm d \ln(P(x))}{\mathrm d \ln(x)}$ mean that $u = \ln(x)$ and we wish to find $\frac{\mathrm d \ln(P(x))}{\mathrm d u}$? –  George V. Williams Jan 21 '13 at 19:52
1  
The notation of the first statement is fine. It means you wish to find the derivative of ln(P(x)) with respect to ln(x). –  daniel.wright Feb 20 '13 at 3:39

1 Answer 1

I have a simpler method:

$\dfrac{d\ln(P(x))}{d\ln(x)}=\dfrac{\dfrac{d\ln(P(x))}{dx}}{\dfrac{d\ln(x)}{dx}}=\dfrac{\dfrac{P'(x)}{P(x)}}{\dfrac{1}{x}}=\dfrac{xP'(x)}{P(x)}$

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That's sneaky! I like it :-D –  Alex Mar 21 '13 at 23:10

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