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Math Help - Differentiation Problem

  1. #1
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    Differentiation Problem

    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?
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  2. #2
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    Re: Differentiation Problem

    Quote Originally Posted by Hank View Post
    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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