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I am having trouble finding the first derivative of $R(P) = (P\textrm{e}^r)(1-\frac{P}{K})$ I am told to use the product rule for this. The first part of it $(P\textrm{e}^r)$ I believe will remain the same for the derivative of it. I am struggling on the $(1-P/K)$.

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The derivative of the function $R(P) = Pe^r\left(1-\frac{P}{K}\right)$ with respect to $P$ can be found by applying the product rule, as follows \begin{align} \frac{d}{dP}R(P) &= \left(\frac{d}{dP}Pe^r\right)\left(1-\frac{P}{K}\right) + Pe^r\frac{d}{dP}\left(1-\frac{P}{K}\right)\\ &= e^r\left(1-\frac{P}{K}\right) + Pe^r\left(\frac{d}{dP}1 + \frac{d}{dP}\left(-\frac{P}{K}\right)\right)\\ &= e^r\left(1-\frac{P}{K}\right) + Pe^r\left(0 - \frac{1}{K}\right)\\ &= e^r\left(1-\frac{P}{K}\right) - \frac{Pe^r}{K}\\ &= e^r\left(1-\frac{2P}{K}\right) \end{align}

A lot of the times, if one does some simple manipulations initially, the amount of work required can be greatly reduced. Notice that $R(P) = Pe^r\left(1-\frac{P}{K}\right) = e^r\left(P-\frac{P^2}{K}\right)$. Now the derivative is simply \begin{align} \frac{d}{dP}R(P) &= e^r\frac{d}{dP}\left(P-\frac{P^2}{K}\right)\\ &= e^r\left(\frac{d}{dP}P-\frac{d}{dP}\frac{P^2}{K}\right)\\ &= e^r\left(1-\frac{2P}{K}\right) \end{align}

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  • $\begingroup$ The origional function Pe^r is being multiplied by (1-P/K) not divided $\endgroup$ Dec 7 '16 at 23:48
  • $\begingroup$ @AmandaSmith Thanks for pointing that out! $\endgroup$
    – Erik M
    Dec 7 '16 at 23:52
  • $\begingroup$ okay that makes alot more sense now! thank you so much $\endgroup$ Dec 8 '16 at 0:20
  • $\begingroup$ what happens to the e^r at the end of e^r(1-P/K)-Pe^r/k $\endgroup$ Dec 8 '16 at 0:30
  • $\begingroup$ @AmandaSmith There's a little intermediate step there, expand it all out: $e^r - Pe^r/K - Pe^r/K = e^r - 2Pe^r/K$ then factor out the $e^r$ in common with both of the terms to get $e^r(1-2P/K)$. $\endgroup$
    – Erik M
    Dec 8 '16 at 0:32
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You may use product rule on $R(P)=e^r(P-\frac{P^2}{K})$ as follows:

$R'(P)=e^r\frac{d}{dP}(P-\frac{P^2}{K})$$+(P-\frac{P^2}{K})\frac{d}{dP}(e^r)$

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