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I have 2 related questions that i think are straightforward but am worried I have over-simplified!

(i) Suppose the demand curve is downward sloping and the supply curve is upwards sloping. Suppose also that I have estimated the price elasticity of demand $ \frac{\partial ln(D)}{\partial ln(p)} $,

Can I obtain the responsiveness of price to demand simply by taking the reciprocal?

$ \dfrac{1}{ \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(p)}{\partial ln(D)} $

By analogy to discrete changes, the rearrangement below seems to work

$ \dfrac{1}{ \frac{\Delta D}{\Delta p} \cdot\frac{p}{D}} = \frac{\Delta p}{\Delta D} \cdot \frac{D}{p} $

Or do I also need an estimate of the price elasticity of supply?

(ii) Relatedly, now suppose I also have the price elasticity of supply, and that I want to estimate the responsiveness of the supply curve to shifts in the demand curve (with the mechanism being that some positive shock to the demand curve raises prices, which incentivises firms to supply more output). Is the following valid?

$ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{ \frac{\partial ln(D)}{\partial ln(p)}} = \frac{\partial ln(S)}{\partial ln(p)} \cdot \frac{\partial ln(p)}{\partial ln(D)} = \frac{\partial ln(S)}{\partial ln(D)} $

Thanks loads for your help, and apologies if I missed a similar question

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    $\begingroup$ Perhaps the following exercise will help you order your thoughts: Suppose $D(p) = 100 - p$, $S(p) = 3p$. Now there is a shift in the demand curve, and it becomes $\hat{D}(p) = 120 - p$. (i) In this case, what do you denote by $\Delta D$, or $\frac{\partial ln(p)}{\partial ln(D)} $? (ii) In this case what do you denote by $\frac{\partial ln(S)}{\partial ln(D)} $? $\endgroup$ – Giskard Feb 22 '19 at 21:47
  • $\begingroup$ In this case, I would have thought (i) $ \Delta D = 20 $ and $ \frac{\partial ln(D)}{\partial ln(p)} = -1 $, thus $ \frac{\partial ln(p)}{\partial ln(D)} = 1/-1 = -1 $. This seems right because if I rearrange the demand curve to get $ p(D) = 100 - D $ and differentiate, I also get $ \frac{\partial ln(p)}{\partial ln(D)} = -1. $ (ii) would be $ 3 * -1 = -3. $ . Is this right? $\endgroup$ – leafnoise Feb 22 '19 at 22:39
  • $\begingroup$ No. Could you please detail your calculations? The numbers you gave seem pretty random. E.g. I see that $\frac{\partial D(p)}{\partial p} = -1$, but why do you think the same is true for $\frac{\partial ln(D)}{\partial ln(p)}$? $\endgroup$ – Giskard Feb 22 '19 at 22:47
  • $\begingroup$ Right, yeah sorry i forgot to convert the differential into an elasticity. So (i) $ \frac{\partial D}{\partial p} \cdot \frac{p}{D} = -1 \cdot \frac{p}{D} = \frac{-p}{D} $. The reciprocal of this is $\frac{-D}{p}$. If i rearrange the demand curve to get $ p(D) = 100 - D$ and find the elasticity that way, i get $ \frac{\partial p}{\partial D} \cdot \frac{D}{p} = -1 \cdot \frac{D}{p} = \frac{-D}{p} $. Then, if my logic for (ii) is right, i would get $ \dfrac{ \frac{\partial ln(S)}{\partial ln(p)}}{\frac{\partial ln(D)}{\partial ln(p)}} = \dfrac{\frac{3p}{S}}{\frac{-p}{D}} = \frac{-3 D}{S}$ $\endgroup$ – leafnoise Feb 23 '19 at 0:02
  • $\begingroup$ "i rearrange the demand curve to get $p(D)=100−D$" This is pretty much the issue, you are denoting both the demand function (a function) and the quantity demanded at a certain price (a number) with $D$. $\endgroup$ – Giskard Feb 23 '19 at 8:32

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