So I'm studying Essential Mathematics for Economic Analysis (Sydsæter, Hammond; page 180) now and stumbled upon this example aimed at illustrating why the Product rule for differentiation works the way it works. I have no problem understanding the formal proof which is given after that example, but the latter drives me crazy.

I would really appreciate if someone helped me with the questions I have. Here's a screenshot from the textbook.

Okay, so we have a formula for the revenue R(P)=P*D(P). Let's say P increases by one dollar and the R changes. That's comprehensible. Why does R increase? Well, R(P) increases by 1*D(P), 'cause each of the D(P) units brings in an extra dollar. But D(P) is also a function and a change in P by one dollar changes the D like: D(P+1)-D(P), which is close to D'(P) That's also more or less clear. But after that I am completely lost.

The (positive) loss due to a one dollar increase in the price per unit is then−P*D'(P), which must be subtracted from D(P) to obtain R'(P), as in equation.

  1. Why do we have have positive loss?
  2. What does it even mean?
  3. And why is this loss -P*D'(P)? Why negative?
  4. And why do we have to subtract it from D(P) to obtain R'(P)? I know that the derivative of R(P)=PD(P) is R'(P)=D(P)+PD'(P). But if I didn't know the derivative why would I subtract this "positive loss" from D(P) to obtain R'(P)?

I hope somebody would be kind enough to answer my ignorant questions. Thanks! :)


Since $D^\prime(P)$ is negative, $-PD^{\prime}(P)$ is positive. Positive loss means the number of dollars lost. For example, if $PD^\prime(P)=-17$ then you lost $17$ dollars (the $17$ is positive, i.e. positive loss). For part 4. you are subtracting the "positive loss" which equates to $-(-PD^{\prime}(P))=+PD^{\prime}(P)$. So the formula is what you expect $$R^{\prime}(P)=D(P)-(-PD^\prime(P))=D(P)+PD^{\prime}(P).$$

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    $\begingroup$ IMO, the explanation in the textbook seems like an overly complicated way to try to make "intuitive" sense of the formula. Of course, the explanation seems to take the intuition out... $\endgroup$ – TravisJ Jun 12 '15 at 13:19
  • $\begingroup$ Thanks a lot! I get it now :) $\endgroup$ – levant Jun 12 '15 at 13:41

Based upon the exposition in the link, here's an alternate (better) viewpoint. Suppose the price $P$ changed by $h$ dollars. Then the revenue from this price change would be given by $$R(P+h) =(P+h)D(P+h).$$ The first factor accounts for the increase in revenue due to price and the second accounts for the change in demand due to the change in price. Note that it is likely the case that $D(P+h) <D(P)$ if the change in price is positive. To see how much the revenue has changed, due to this price change, subtract $R(P)$ from $R(P+h)$. This gives $$R(P+h)-R(P) = (P+h)D(P+h)-PD(P)=hD(P+h)+P[D(P+h)-D(P)].$$ Now to see the rate at which revenue is changing, due to the change in price $h$, divide both sides of the above by $h$. This gives $$\frac{R(P+h)-R(P)}{h} = D(P+h)+P\frac{D(P+h)-D(P)}{h}.$$ Note that for small changes in price, the above equation is approximately $$R^{\prime}(P)=D(P)+PD^{\prime}(P).$$ This is the product rule.

  • $\begingroup$ Thank you. I found your explanation much clearer than the one in the textbook. There's something I didn't get though (sorry, I am thickheaded). Would you please be so kind to explain, why do you multiply the right-hand side of your equation by h? It's perfectly clear to me that R(P+h)−R(P)=(P+h)D(P+h)−PD(P). But then you have h out of nowhere... $\endgroup$ – levant Jun 12 '15 at 14:42
  • $\begingroup$ Take the formula you have written as distribute the $P+h$ factor in the first term on the LHS. $\endgroup$ – Lythia Jun 12 '15 at 16:39

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