Re: Differentiation Problem

Quote:

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.

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