What is the derivative of t^3/3
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
3t^2
don't forget that 1/3 up on the first line
t^2
the end
I do not want the limit i just need the derivative so I can set it equal to 15
That is the derivative.
d/dt (t^3/3) = t^2
Since you asked for the derivative, I derived the derivative.
Well can you make it easier to understand how you got it?
A derivative IS a limit !
I mean how you got it? Your explanation is very confusing
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
or
derivative of (1/3)t^3 = t^2
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
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.
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