<p>You're correct. Many useful things can be shown by taking this derivative.</p>
<p>Perhaps you were confused because you are used to solving for quantity rather than price - in which case you are more likely to take the derivative of R = z * p(z) with respect to z which gives you p(z) + z * dp(z)/dz</p>

CC BY-SA 4.0
2020-08-27T18:36:48.270Z
2020-08-27T18:36:48.270Z
Dek
<p><span class="math-container">$p\cdot z(p)$</span> is total revenue (price times quantity), and so its derivative <span class="math-container">$\frac{\mathrm d}{\mathrm dp}pz(p)=pz'(p)+z(p)$</span> is marginal revenue.</p>

CC BY-SA 4.0
2020-08-25T17:21:29.387Z
2020-08-25T17:21:29.387Z
Herr K.
<p>Suppose that <span class="math-container">$z(\cdot)$</span> is the demand function of an individual (consumer/investor) and <span class="math-container">$p$</span> is the price of the commodity/asset demanded. Does anybody know what is the intuition behind the following expression?
<span class="math-container">\begin{align}pz^{'}(p)+z(p)\end{align}</span></p>
<p>It seems to me like the derivative of <span class="math-container">$(pz(p))^{'}$</span> with respect to <span class="math-container">$p$</span>. What is the intuition behind this? Has anybody seen anything this expression before?</p>

CC BY-SA 4.0
2020-08-25T16:49:38.223Z
Marginal revenue
Herr K.
CC BY-SA 4.0
2020-08-25T16:57:34.080Z
Thank you @Herr K. My mind has stuck! You are absolutely right!
Hunger Learn
<microeconomics><mathematical-economics><finance>
Has anybody seen anything this expression before?
Hunger Learn
# Has anybody seen anything this expression before?

(Asked by: username_0 on 2020-08-25)

Suppose that $z(\cdot)$ is the demand function of an individual (consumer/investor) and $p$ is the price of the commodity/asset demanded. Does anybody know what is the intuition behind the following expression?
\begin{align}pz^{'}(p)+z(p)\end{align}

It seems to me like the derivative of $(pz(p))^{'}$ with respect to $p$. What is the intuition behind this? Has anybody seen anything this expression before?

**Tags:** microeconomics, mathematical-economics, finance

> **Comments:**
> **username_1**: Marginal revenue  
> **username_0**: Thank you @username_1 My mind has stuck! You are absolutely right!  

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## 2 answers

### Answer by username_1 on 2020-08-25. Score: 9 | Accepted answer

$p\cdot z(p)$ is total revenue (price times quantity), and so its derivative $\frac{\mathrm d}{\mathrm dp}pz(p)=pz'(p)+z(p)$ is marginal revenue.


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### Answer by username_2 on 2020-08-27. Score: 2

You're correct. Many useful things can be shown by taking this derivative.

Perhaps you were confused because you are used to solving for quantity rather than price - in which case you are more likely to take the derivative of R = z \* p(z) with respect to z which gives you p(z) + z \* dp(z)/dz


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