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For the following relation

$\log (\mathrm{Q})=4.415-5.132 \times \log (\mathrm{P})+\mathrm{e}$

I need to prove that:

$$\frac{\mathrm{d} \log (\mathrm{Q})}{\mathrm{d} \log (\mathrm{P})}=\frac{\dfrac{\mathrm{dQ}}{\mathrm{Q}}}{\dfrac{\mathrm{dP}}{\mathrm{P}}} $$

I tried to use the chain rule, since we see that $\log(Q)$ is actually a function that can be written as $f(Q(P))$. Unfortunately I did not succeed in writing the proof. The problem is that I can solve $\frac{d\log(Q)}{dP} = \frac{d\log(Q)}{dQ} \frac{dQ}{dP}$, but the $d\log(P)$ term is nowhere to be found in this expression.

Could I please get feedback on how to approach this problem?

Sebastiano
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Tim
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1 Answers1

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This is not Leibniz notation such as what you used in your chain rule: you have, for example, $\dfrac{\mathrm dQ}{Q}$, which is $\dfrac1Q \mathrm dQ$ where $\mathrm dQ$ is a differential.

The calculation is simply

$$ \mathrm d \log(Q) = \frac{\mathrm d \log(Q)}{\mathrm dQ} \mathrm dQ = \dfrac 1Q \mathrm dQ $$

where $\dfrac{\mathrm d \log(Q)}{\mathrm dQ}$ is the derivative of $\log(Q)$ with respect to $Q$, that is, this part of the equation is a Leibniz notation.

Do that kind of calculation for both top and bottom of the original $\dfrac{\mathrm d \log(Q)}{\mathrm d \log(P)}$ and you end up with

$$ \frac{\dfrac 1Q \mathrm dQ}{\dfrac 1P \mathrm dP} $$

which is written a bit more picturesquely as

$$ \frac{\dfrac{\mathrm dQ}Q}{\dfrac{\mathrm dP}P}. $$

David K
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  • Amazing! Thanks for the clear explaination. Due to a lack of knowledge it's not entirely clear why the (dlog(Q)/dQ) term is multiplied with a dQ term. Could you as a final remark shed a light upon that? – Tim Jan 07 '23 at 23:26
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    The equation is describing the relationship between two differentials, $\mathrm d\log(Q)$ and $\mathrm dQ.$ There has to be a differential on each side so that you have two comparable objects. The term $\frac{\mathrm d \log(Q)}{\mathrm dQ}$ is just a factor that accounts for how different the magnitudes of the two differentials are. So rather than say $\frac{\mathrm d \log(Q)}{\mathrm dQ}$ is a term that is multiplied by $\mathrm dQ$, I think it makes more sense to say that $\mathrm dQ$ has to be multiplied by $\frac{\mathrm d \log(Q)}{\mathrm dQ}$. – David K Jan 07 '23 at 23:55
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    It might help to think of this as like what you do for U-substitution in an integral: $\mathrm du = u'(x),\mathrm dx$. – David K Jan 07 '23 at 23:58