Question

Use the definition of the derivative to find f'(t) where f(t) = (t^2-3).

Answer 1

Use this and replace f(x) with f(t), which is (t^2-3).

Answer 2

Can you help set me up with a few steps? I keep canceling everything out and I know that can't be right.

Answer 3

Oh, sure. Let me work it out first.

Answer 4

\[f'(t) = \lim(h->0) \frac{ f(t+h) - f(t)}{ h }\]
\[f'(t) = \lim(h->0) \frac{ [(t+h)^{2}-3] - [(t^{2}-3)]}{ h }\]

Answer 5

Let me know if you'd like me to continue.

Answer 6

yes, the next step is where I am getting confused can you please continue

Answer 7

\[f'(t) = \lim(h->0) \frac{ f(t+h) - f(t)}{ h }\]
\[f'(t) = \lim(h->0) \frac{ [(t+h)^{2}-3] - [(t^{2}-3)]}{ h }\]
\[f'(t) = \lim(h->0) \frac{ (t^{2}+2th+h^{2}-3) - (t^{2}-3)}{ h }\]

Answer 8

So, in the next step I can cancel out the 3 and the t^2 and should be left with (2th+h^(2))/h. Then I factor our the h on top and that cancels out the denominator and then I am left with 2t+h and since the limit goes to 0 the h gets cancelled and I am left with 2t and the answer?

Answer 9

Yes! :D

Answer 10

Got the same answer.