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Finding first and second derivative

  1. Feb 12, 2009 #1
    how would I find the first and second derivative of

    (x^3/3) - 2x^2 + 3x +8

    thanks
     
  2. jcsd
  3. Feb 12, 2009 #2

    Mark44

    Staff: Mentor

    By differentiating (x^3/3) - 2x^2 + 3x +8,
    and then differentiating what you got from the previous step.

    What have you tried?
     
  4. Feb 12, 2009 #3
    thanks for the response, I am just having trouble with the understanding the fraction (x^3/3) and how to find the derivative. I can do the rest fine if I can figure out how to get the derivative of (x^3/3).

    could I do this, multiply the whole equation by 3 to get rid of the fraction? Can someone be kind enough to show me how to find the derivative of a fraction with an x variable int he numerator? I know something like 3/x^3 is rewritable as 3^-3 but that has the x in the denominator.

    thanks
     
  5. Feb 12, 2009 #4

    Mark44

    Staff: Mentor

    You don't really have an equation, so you can't multiply it by 3.

    x^3/3 is the same as 1/3 * x^3. Do you know that d/dx(a*f(x)) = a * d/dx(f(x))?

    BTW, 3/(x^3) = 3*x^(-3)
     
  6. Feb 12, 2009 #5
    yes I understand how and have found the derivative of (x^3/3) is x^2 but don't understand how exactly one gets to that. Do you just take the derivative of x^3 and get x^2 and then derivative of 3 which is 0 so you get x^2/0 = 0? thats what I was trying to do but don't see how you get x^2, do you just ignore the constant?

    thanks
     
  7. Feb 12, 2009 #6
    If you just follow the formula (d/dx)x^n=n*x^(n-1). So for example (1/3)*x^3.
    (d/dx)(1/3)*x^3
    =(1/3)*3*x^2 (the constant can be pulled out front)
    = x^2
     
  8. Feb 12, 2009 #7
    thanks, this is actually a rectilinear motion problem

    alright, now the next step, original problem is

    f(x) = x^3/3-2x^2 + 3x +8

    first derivative: x^2 - 4x + 3 .... x = 3 x = 1
    second derivative: 2x - 4 .... x = 2

    s(0) = 8
    s(1) = 9 1/3
    s(2) = 8 2/3
    s(3) = 8

    so now

    time | velocity | acceleration | speed
    0 < x <1
    1< x < 2
    2< x < 3
    x> 3

    how would I find out the velocity acceleration speed?
     
  9. Feb 12, 2009 #8

    Mark44

    Staff: Mentor

    You have too much stuff.
    f'(x) = x^2 - 4x + 3 PERIOD
    f''(x) = 2x - 4 PERIOD

    Now if you wanted to solve the equation f'(x) = 0, which you said nothing about, you would get x = 3 or x = 1. But the first derivative, or f'(x) is just x^2 - 4x + 3.

    Similarly for the second derivative: f''(x) = 2x -4.
    You didn't identify a function with a letter in your first post, but it now appears that what you started with was s(t) = 1/3 * x^3 - 2x^2 + 3x + 8.

    The first and second derivatives, s'(t) and s''(t) are what you found.
    If s(t) is the position at time t, s'(t) is the velocity, and s''(t) is the acceleration. The speed is |s'(t)|.

    Apparently you have to fill out a table, but I don't know what's supposed to go into the table. You can't put an interval (e.g., 0 < x < 1) into a function.
     
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