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. 
 A: $$\begin{align*}
\lim_{h\to 0}\frac{f(t+h) - f(t)}{h} &= \lim_{h\to 0}\frac{\frac{2(t+h)+1}{(t+h)+3} - \frac{2t+1}{t+3}}{h}
\\
&\strut\\
&=\lim_{h\to 0}\frac{\frac{2t+2h+1}{t+h+3}- \frac{2t+1}{t+3}}{\frac{h}{1}}\\
&\strut\\
&= \lim_{h\to 0}\frac{1}{h}\left(\frac{2t+2h+1}{t+h+3} - \frac{2t+1}{t+3}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3)}{(t+h+3)(t+3)} - \frac{(2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3) - (2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + t + 6t + 6h + 3 - (2t^2 + 2th + 6t +t + h + 3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + 7t + 6h + 3 - 2t^2 - 2th - 7t -h - 3}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{5h}{(t+3)(t+h+3)}\right).
\end{align*}$$
Nothing but algebra so far. 
Can you take it from here?
If you made a mistake in your own derivation, have you spotted it? 
A: 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.
A: $\frac{(2t+1)}{(t+3)}$
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 $h(x)$
$f'(x) = \frac{h(x)g\prime(x)-h\prime(x)g(x)}{[h(x)^2]}$
$f'(x) = \frac{(t+3)2-1(2t+1)}{(t+3)^2}$
$f'(x) = \frac{2t+6-2t-1}{(t+3)^2}$
$f'(x) = \frac{5}{(t+3)^2}$
A: $f( t)=\frac{{2t + 1}}{{t + 3}}=\frac{{2t + 6 - 5}}{{t + 3}} = \frac{{2(t + 3) - 5}}{{t + 3}} = \frac{{2(t + 3)}}{{t + 3}} - \frac{5}{{t + 3}} = 2 - \frac{5}{{t + 3}}$
$\text{Let } g(t) = \frac{5}{{t + 3}}{\text{ , so  }}f(t) = 2 - g(t) \Rightarrow f'(t) = 2' - g'(t) \Rightarrow f'(t) = -g'(t)$
$\text{Since }(\frac{u}{v})' = \frac{{u'v - v'u}}{{{v^2}}} \Rightarrow g'(t) = \frac{{5'(t + 3) - 5(t + 3)'}}{{{{(t + 3)}^2}}} \Rightarrow g'(t) =  - \frac{{5(t' + 3')}}{{{{(t + 3)}^2}}}$
$g'(t) =  - \frac{5}{{{{(t + 3)}^2}}}\text{ ,and since we showed that }f'(t) =- g'(t) \Rightarrow f'(t) = \frac{5}{{{{(t + 3)}^2}}}$
