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I can't figure out where I erred in calculating this derivative:

$$f(t) = 4.9t^2$$

$$f'(t) = \frac {f(t+h) - f(t)} {h} = \frac {4.9(t+h)^2 - 4.9t^2} {h}$$ $$= \frac {4.9(t^2 + h^2 +2ht) - 4.9t^2} {h} = \frac {4.9t^2 + 4.9h ^2 + 9.8ht - 4.9t^2} { h}$$ $$=\frac {4.9h^2 + 9.8ht} { h} = 4.9h+9.8t$$

Obviously the correct answer is $9.8t$. I think my fault was in the last step, for some reason I'm still a bit confused when it comes to dividing the difference by $h$.

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    Your definition of the derivative is incorrect. $f'(t)$ does not equal $\frac{f(t+h)-f(t)}{h}$ but the limit $\displaystyle\lim_{h\rightarrow 0}\frac{f(t+h)-f(t)}{h}.$ Thus taking the limit of $4.9h +9.8t$ as $h$ goes to $0$ gives you the correct answer.2011-09-20

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Your error was simply that $$f'(t)=\lim_{h\to 0}\frac{f(t+h)-f(t)}{h},$$ not $$f'(t)=\frac{f(t+h)-f(t)}{h}.$$ Take the limit of the expression you got, $ 4.9h+9.8t$, as $h\to 0$, and you get the correct answer :)

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    Cool. So if its was $4.9(2h)$ or $4.9h^2$, those would likewise evaluate to $0$?2011-09-20
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    @Matt: Yup!$\text{}$2011-09-20