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Thread: Delta-Y/Delta-X as a function of X and Delta-X (econ related

  1. #1

    Delta-Y/Delta-X as a function of X and Delta-X (econ related

    I've got the following review problem in my micro-econ course:

    y = f(x) = 5x^2 - 4x
    (a) Find delta-y/delta-x as a function of x and delta-x.
    (b) Find dy/dx.

    (b)'s just the derivative and I've confirmed the answer in the book. However, I cannot work out how one arrives at this solution for (a): 10x + (5delta-x) - 4

    Where does the 5delta-x come from?

    The text gives this form:
    delta-y/delta-x = [f(x - delta-x) - f(x)] / delta-x

    Using that, I got 5delta-x - 6

    Any help would be greatly appreciated. Thanks in advance.

  2. #2
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    Re: Delta-Y/Delta-X as a function of X and Delta-X (econ rel

    [tex]f(x ) \ = \ 5x^2-4x, \ \frac{dy}{dx} \ = \ \frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

    [tex]= \ \frac{5(x+\Delta x)^2-4(x+\Delta x)- (5x^2-4x)}{\Delta x}[/tex]

    [tex]= \ \frac{5(x^2+2x\Delta x+\Delta x^2)-4x-4\Delta x-5x^2+4x}{\Delta x}[/tex]

    [tex]= \ \frac{5x^2+10x\Delta x+5\Delta x^2-4x-4\Delta x-5x^2+4x}{\Delta x}[/tex]

    [tex]= \ \frac{10x\Delta x+5\Delta x^2-4\Delta x}{\Delta x}[/tex]

    [tex]= \ 10x+5\Delta x-4[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

  3. #3

    Re: Delta-Y/Delta-X as a function of X and Delta-X (econ rel

    Thanks for the steps!

    Looks like I could've benefited from having worked it a third time. I now see what I'd been doing incorrectly.

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