Differentiation Problem

**Hank** · September 17th 2012, 08:50 AM

If p is the price per unit at which x units of a commodity can be sold then R=xp is called the total revenue and dR/dx is the marginal revenue. Show that the marginal revenue is p + x dp/dx.

I found that the derivative of R is P because
R + dR = (x + dx)p
R + dR = xp + (dx)p
(subtracting the orginal equation R = xp)
dR = (dx)p
(divided by delta x)
dR/dx = p

So basically I need to show that dp/dx = 0 because then
dR/dx = p + x dp/dx
dR/dx = p + x(0)
dR/dx = p

The problem is when I differentiate dp/dx I get this
P=R/x
P + dp= R/(x+dx)
P + dp= R/x + R/dx
dp = R/dx
dp/dx = (R/dx)/dx

Or if I find the derivative by nx^n-1 I get dp/dx = R

Neither of these values of dp/dx make x(dp/dx) equal 0, what am I doing wrong?

**Prove It** · September 17th 2012, 08:56 AM

Originally Posted by Hank

If p is the price per unit at which x units of a commodity can be sold then R=xp is called the total revenue and dR/dx is the marginal revenue. Show that the marginal revenue is p + x dp/dx.

I found that the derivative of R is P because
R + dR = (x + dx)p
R + dR = xp + (dx)p
(subtracting the orginal equation R = xp)
dR = (dx)p
(divided by delta x)
dR/dx = p

So basically I need to show that dp/dx = 0 because then
dR/dx = p + x dp/dx
dR/dx = p + x(0)
dR/dx = p

The problem is when I differentiate dp/dx I get this
P=R/x
P + dp= R/(x+dx)
P + dp= R/x + R/dx
dp = R/dx
dp/dx = (R/dx)/dx

Or if I find the derivative by nx^n-1 I get dp/dx = R

Neither of these values of dp/dx make x(dp/dx) equal 0, what am I doing wrong?

This is a simple application of implicit differentiation.

$\displaystyle \begin{align*} R &= P\,x \\ \frac{d}{dx}\left( R \right) &= \frac{d}{dx}\left( P\,x \right) \\ \frac{dR}{dx} &= P\,\frac{d}{dx}\left(x\right) + x\,\frac{d}{dx}\left(P\right) \\ \frac{dR}{dx} &= P\cdot 1 + x\,\frac{dP}{dx} \\ \frac{dR}{dx} &= P + x\,\frac{dP}{dx} \end{align*}$

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