What is the derivative of t^3/3

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asked by Britt
  1. limit of (1/3)[(t+h)^3 -t^3]/h as h--> 0
    (t+h)^3 = t^3 + 3t^2h + 3th^2 + h^3
    subtract t^3
    3t^2h + 3 th^2 + h^3
    divide by ha
    3t^2 +3th +h^2
    let h --> 0
    don't forget that 1/3 up on the first line
    the end

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    posted by Damon
  2. I do not want the limit i just need the derivative so I can set it equal to 15

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    posted by Britt
  3. That is the derivative.
    d/dt (t^3/3) = t^2

    Since you asked for the derivative, I derived the derivative.

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    posted by Damon
  4. Well can you make it easier to understand how you got it?

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    posted by Britt
  5. A derivative IS a limit !

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    posted by Damon
  6. I mean how you got it? Your explanation is very confusing

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    posted by Britt
  7. In general

    Look at the tangent to f(x) at x

    now look at the function a little bit further, at x + some little h

    if the curve is smooth, the tangent goes close to the point
    [(x+h), f(x+h) ]
    as the f(x) goes up or down between x and x+h

    now as h gets tiny, the slope gets closer and closer to
    [f(x+h)-f(x)] / [(x+h)-h]
    that denominator is h of course
    so the slope (derivative) approaches
    [f(x+h)-f(x)] / h as h gets tiny (goes to zero)

    That is the definition of the derivative that I used to get
    d/dt (t^3) = 3 t^2
    derivative of (1/3)t^3 = t^2

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    posted by Damon
  8. Look in the index of your calculus book for derivative, definition of derivative. They may explain it better or search on google for "definition of derivative" calculus

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    posted by Damon
  9. When I did that search, one of the first results was a " Wikipedia" article that does what I did with very nice graphs and stuff so it may be easier to understand than my trying to type it freehand.

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    posted by Damon
  10. This is much simpler... (t^3)/3

    Write as 1/3(t^3) so that the 1/3 is multiplied rather than dividing by 3.

    Using the power rule, which states that the derivative of x^n is n*x^(n-1), we get...

    f'(x) = (3)(1/3)(t^2) = t^2

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    posted by Michael
  11. That is completely correct Michael, but Britt asked where it came from. That is why I started with the definition.

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    posted by Damon

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