# Finding $f'(t)$ for a quotient

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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Try $f(t)=2-5/(t+3)$ and proceed. – Did Sep 11 '11 at 22:00
Where does 2-5 come from? – user138246 Sep 11 '11 at 22:01
Try to compute $f(t)-2$ and the light will come. – Did Sep 11 '11 at 22:02
@Jordan: Long division:$$\frac{2t+1}{t+3} = \frac{2(t+3)-5}{t+3} = \frac{2(t+3)}{t+3}+\frac{-5}{t+3} = 2 - \frac{5}{t+3}.$$ – Arturo Magidin Sep 11 '11 at 22:03
I am not sure what that means but I got $(2(a+h)+1) /(a+h+3)$ – user138246 Sep 11 '11 at 22:04

\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: What "first step all over $t+3$"? In the first step we just have, inside the limit,$$\frac{f(t+h)-f(t)}{h} = \frac{\quad\frac{2(t+h)+1}{(t+h)+3} - \frac{2t+1}{t+3}\quad}{h}.$$Just plug in $t+h$ for $t$ into the formula for $f$ to find $f(t+h)$. Not everything is divided by $t+3$. – Arturo Magidin Sep 11 '11 at 22:19
Okay I think I follow, why did you use 1/h in the problem though, is that just to make it look cleaner? – user138246 Sep 11 '11 at 22:23
I think I got it all, I just don't know what to do next. – user138246 Sep 11 '11 at 22:25
@Jordan: Dividing by $h$ is the same thing as multiplying by $\frac{1}{h}$. Because it is easy to get confused when you have fractions of fractions, it is best to turn it into simple fractions quickly, to avoid errors down the line. So instead of having a fraction divided by $h$, we have a fraction multiplied by $\frac{1}{h}$. – Arturo Magidin Sep 11 '11 at 22:26
@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 $0$? – Arturo Magidin Sep 11 '11 at 22:26

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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I have no idea about any of these rules. I just looked in my book, it is in the next chapter. – user138246 Sep 11 '11 at 22:06
@Jordan: Oh, that's too bad. Well, then Arturo and Didier have the right path, and that's what you need to do. – mixedmath Sep 11 '11 at 22:09

$\frac{(2t+1)}{(t+3)}$

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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Hmm, out of curiosity what is wrong with this? – Dair Sep 11 '11 at 22:15
I do not know the quotient rules. That is not until my next chapter. – user138246 Sep 11 '11 at 22:16
@Jordan: Are you not getting this from the chapter with the quotient rule? – Dair Sep 11 '11 at 22:21
I am using the horrible Early Transcendental 7e Steward book. I have a test on chapter 2, but the rules are in chapter 3. – user138246 Sep 11 '11 at 22:26

$f( t)=\frac{{2t + 1}}{{t + 3}}=\frac{{2t + 6 - 5}}{{t + 3}} = \frac{{2(t + 3) - 5}}{{t + 3}} = \frac{{2(t + 3)}}{{t + 3}} - \frac{5}{{t + 3}} = 2 - \frac{5}{{t + 3}}$

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

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

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

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