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Question about derivatives.

  1. Jun 12, 2008 #1
    Hello guys

    I've been recently thinking about derivative and it's applications.

    Let's say Y=3x

    When X = 1, Y=3
    wHEN X = 2, Y = 6
    When X = 3, Y = 9

    Dy/Dx is the rate of change in Y with respect X, it is 6-3/2-1 =3/1 = 3
    and the derivative is 3.

    Let's Y = (X)^2

    When X = 1, Y=1
    When X = 2, Y = 4
    When X = 3 , Y=9
    When X = 4 , Y = 16

    Dy/dx = 4-1/2-1 = 3/1 = 3
    Dy/dx = 9-4/3-2 =5/1 = 5
    Dy Dx = 16-9/4-3=7/1 = 7

    So the rate of Change is not constant.

    When you derive you get 2x

    When you plug values in the function, you get different values from the the rate of change, Where is my mistake?

  2. jcsd
  3. Jun 12, 2008 #2


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    The function is different than its derivative. Of course, the values at any given x can be different.
  4. Jun 12, 2008 #3


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    Hi racer! :smile:

    You have to choose the half-way value …

    3 5 and 7 correspond to 1,2 2,3 and 3,4.

    half-way is 1.5 2.5 and 3.5, and 2x of that is 3 5 and 7 !

    Happier? :smile:
  5. Jun 12, 2008 #4


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    You should expect dy/dx to be different than (y(x2)-y(x1)) / (x2-x1). The former is exactly the rate of change, while the latter is just an approximation to the rate of change.
  6. Jun 12, 2008 #5


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    Much what the others are saying but:

    [f(x1)- f(x0)]/(x1- x0) is the average rate of change of f between x0 and x1. The derivative of f (at a point x= a) is the instantaneous rate of change at x= a. If a function is linear (as your example f(x)= 3x) then the average rate of change over interval, starting at any point is a constant (the slope of the line) so the two are exactly the same. If a function is not linear (as f(x)= x2), the average rate of change depends upon both the starting point and the interval and the "average rate of change" and derivative are different.
  7. Jun 13, 2008 #6
    That was amazing, I'll see if this applies to other functions.

    Is there a realistic proof or example that contains releastic things?

    It is logical, I would like to see a realistic example and I know that it is hard to find a realistic example that proves derivatives but there has to be an example proves derivatives realistically specially of functions that are second degree and above.

    Yeah, I made a mistake because I read somewhere that the tan of the angle between the line that a function makes and the x Axis is the first derivative.

    Thanks guys.
  8. Jun 13, 2008 #7
    Please note the difference:

    The slope formula will give you the rate of change between two points.

    The derivative function will give you the rate of change at one particular point.

    That's a big difference! Hopefully you will take some time to think that through.

    So say you have any function f(x). Pick two points [tex](x_1, y_1) [/tex] and [tex](x_2, y_2)[/tex]

    Now from what I have said above, you should know tha the slope formula will allow you to calculate the rate of change between these two points. Where as the derivative, f'(x), give you the rate of change at one particular point.

    One interesting result, if you know the rate of change between two points (call it m), you are not guaranteed that the rate of change at either of those points is m! However, you are guaranteed that there is a point between those two points where the rate of change is m! In the example of f(x) = x^2, it just so happens that the point where this occurs is half way between any two points. For other functions it probably won't be exactly halfway.
    Last edited: Jun 13, 2008
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