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Rate of change in a function?

  1. Nov 25, 2009 #1
    Is differentiation just finding the rate of change in a function?

    [tex]\frac{dx}{dy}[/tex] (y=6x) =6?


    rate of change in [tex]\frac{2}{y}=\frac{2}{x}[/tex]the same as :[tex]\frac{dx/dy}{dy/dx}*y=\frac{5y}{2}[/tex] if and only if y=2/5 ? which equals 1 lol

    I think I am right :x but non- of the Wikipedia articles help because of there complexity.
     
    Last edited: Nov 25, 2009
  2. jcsd
  3. Nov 25, 2009 #2
    Re: differentiation

    Differentiation is with respect to some function, dy/dx is just a notation used to denote the derivative, in the case of dy/dx it denotes the derivative of the function y=f(x) (y as a function of x). dx/dy should denote the derivative of the function x=f(y), x as a function of y, in the first case x = y/6 => dx/dy = 1/6.

    If 2/y = 2/x , you can solve for the variable of interest, i.e. y=x, but this definition only holds for x not equal to 0 in which case 2/y = 2/x does not tell us how y depends on x because 2/x is undefined.
     
  4. Nov 26, 2009 #3
    Re: differentiation

    considering I have no clue what you just said:

    "The derivative of a function represents an infinitesimal change in the function with respect to one of its variables. "


    Does this mean what I said before?
     
  5. Nov 26, 2009 #4

    HallsofIvy

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    Re: differentiation

    The derivative of a function is its instantaneous rate of change rather than its average rate of change. For x from 0 to 4, [itex]x^3[/itex] changes from 0 to 64 so has an average rate of change of 64/4= 8. For x from 1 to 3, [itex]x^3[/itex] changes from 1 to 27 and so has an average rate of change of (27- 1)/(3- 1)= 26/2= 13. From 1.5 to 2.5, [itex]x^3[/itex] changes from 3.375 to 15.625 and so has an average rate of change of (15.625- 3.375)/(2.5-1.5)= 12.25/1= 12.25.

    According to calculus, the derivative of [itex]x^3[/itex] is [itex]3x^2[/itex] and, at x= 2, that is 3(4)= 12. That is rate of change at x= 2.

    By the way, what you had initially,
    [tex]\frac{dx}{dy}[/tex](y= 6)= 6.
    makes no sense. I think you mean to ask about the derivative of the function 6x but that is not what you wrote. First, "y= 6x" is an equation, not a function. Second, the derivative of a function, say, f(x), is written as "df(x)/dx", NOT "dy/dx(f(x)". Oh, and you had "y" and "x" reversed. You wanted "d(6x)/dx" or "dy/dx" with y= 6x.

    And, yes, a linear function has the property that its rate of change is the same at every point so "average rate of change" and "instanteous rate of change", that is the "rate of change at a particular x" is that same constant- the slope of the line graph.
     
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