# Either find an example of a differentiable downward-sloping function $p$ such that $\frac{d}{dx}[p(x)+xp'(x)]<0$

In economic models we assume that firms which face a decreasing demand curve also face a decreasing marginal revenue curve. I just realized that I wasn't sure that this is true, so I tried to prove it, however I'm finding this difficult.

Let $p(x)$ be an inverse demand curve such that $p'(x)<0$ (so, if you're unfamiliar with demand curves and things, $p(x)$ is the price that would elicit consumers to buy $x$ units of a good). Then total revenue is given by $xp(x)$, and so the marginal revenue, which is just the derivative of total revenue, would be $\frac{d}{dx}(xp(x))=p(x)+xp'(x)$. I would like, from the assumption that $p'(x)<0$, to show that $\frac{d}{dx}[p(x)+xp'(x)]<0$... but I suspect from looking at it that it's not even necessarily true. Taking the derivative of that gives that it's true as long as $2p'(x)<xp''(x)$, but that's not very meaningful to me. Can you see from that what sorts of functions might violate that condition, or if it's always true?

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The question in the title doesn't match the question in the body. Please edit to bring the two into agreement. Also, what's $q$? – Gerry Myerson Oct 21 '12 at 23:35
@GerryMyerson I think I've fixed it... $q$ was a typo – crf Oct 21 '12 at 23:40

Let $p(x)=e^{-x}$. Then $p'(x)=-e^{-x}\lt0$ for all $x$. $xp(x)=xe^{-x}$. $(xp(x))'=p(x)+xp'(x)=-xe^{-x}+e^{-x}=(1-x)e^{-x}$. $(p(x)+xp'(x))'=(x-2)e^{-x}$. This is positive for $x\gt2$.