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I am suppose to find f'(t) for $f(t)=\frac{2t+1}{t+3} .$

I know it seems simple, I plug everything in and I don't get close to the proper answer.

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    @Jordan: Or as in doing what I did above algebraically; or as in doing that division in your head. But yes, basically that.2011-09-11

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$\begin{align*} \lim_{h\to 0}\frac{f(t+h) - f(t)}{h} &= \lim_{h\to 0}\frac{\frac{2(t+h)+1}{(t+h)+3} - \frac{2t+1}{t+3}}{h} \\ &\strut\\ &=\lim_{h\to 0}\frac{\frac{2t+2h+1}{t+h+3}- \frac{2t+1}{t+3}}{\frac{h}{1}}\\ &\strut\\ &= \lim_{h\to 0}\frac{1}{h}\left(\frac{2t+2h+1}{t+h+3} - \frac{2t+1}{t+3}\right)\\ &\strut\\ &=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3)}{(t+h+3)(t+3)} - \frac{(2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\ &\strut\\ &=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3) - (2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\ &\strut\\ &=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + t + 6t + 6h + 3 - (2t^2 + 2th + 6t +t + h + 3)}{(t+3)(t+h+3)}\right)\\ &\strut\\ &=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + 7t + 6h + 3 - 2t^2 - 2th - 7t -h - 3}{(t+3)(t+h+3)}\right)\\ &\strut\\ &=\lim_{h\to 0}\frac{1}{h}\left(\frac{5h}{(t+3)(t+h+3)}\right). \end{align*}$ Nothing but algebra so far.

Can you take it from here?

If you made a mistake in your own derivation, have you spotted it?

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    @Jordan: Same as usual: If you try evaluating you get $\frac{0}{0}$. Is there something easy to cancel in the expression we have, so that the resulting limit can be computed by simply plugging in $0$?2011-09-11
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While people tout the quotient rule, I want to mention that performing the product rule can work here too. Usually, the product rule says that (f(x)g(x))' = f'(x)g(x) + f(x)g'(x), so here you can take $f(t) = (2t+1)$ and $g(t) = \dfrac{1}{t+3} = (t+3)^{-1}$.

I'm assuming you know the quotient rule and it's just not coming out right, so I'm giving you a method to check those answers. Of course, you could also check your product rule derivatives with the quotient rule in the same way, but that sounds unwieldy.

I also want to remind you to remember the chain rule.

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    @Jordan: Oh, that's too bad. Well, then Arturo and Didier have the right path, and that's what you need to do.2011-09-11
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$\frac{(2t+1)}{(t+3)}$

You can use the quotient rule:

http://en.wikipedia.org/wiki/Quotient_rule

Make $2t+1$ the function $g(x)$ and make $(t+3)$ the function $h(x)$

f'(x) = \frac{h(x)g\prime(x)-h\prime(x)g(x)}{[h(x)^2]}

f'(x) = \frac{(t+3)2-1(2t+1)}{(t+3)^2}

f'(x) = \frac{2t+6-2t-1}{(t+3)^2}

f'(x) = \frac{5}{(t+3)^2}

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    I am using the horrible Early Transcendental 7e Steward book. I have a test on chapter 2, but the rules are in chapter 3.2011-09-11
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$f( t)=\frac{{2t + 1}}{{t + 3}}=\frac{{2t + 6 - 5}}{{t + 3}} = \frac{{2(t + 3) - 5}}{{t + 3}} = \frac{{2(t + 3)}}{{t + 3}} - \frac{5}{{t + 3}} = 2 - \frac{5}{{t + 3}}$

\text{Let } g(t) = \frac{5}{{t + 3}}{\text{ , so }}f(t) = 2 - g(t) \Rightarrow f'(t) = 2' - g'(t) \Rightarrow f'(t) = -g'(t)

\text{Since }(\frac{u}{v})' = \frac{{u'v - v'u}}{{{v^2}}} \Rightarrow g'(t) = \frac{{5'(t + 3) - 5(t + 3)'}}{{{{(t + 3)}^2}}} \Rightarrow g'(t) = - \frac{{5(t' + 3')}}{{{{(t + 3)}^2}}}

g'(t) = - \frac{5}{{{{(t + 3)}^2}}}\text{ ,and since we showed that }f'(t) =- g'(t) \Rightarrow f'(t) = \frac{5}{{{{(t + 3)}^2}}}