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. 

$$\begin{align*}
\lim_{h\to 0}\frac{f(t+h) - f(t)}{h} &= \lim_{h\to 0}\frac{\frac{2(t+h)+1}{(t+h)+3} - \frac{2t+1}{t+3}}{h}
\\
&\strut\\
&=\lim_{h\to 0}\frac{\frac{2t+2h+1}{t+h+3}- \frac{2t+1}{t+3}}{\frac{h}{1}}\\
&\strut\\
&= \lim_{h\to 0}\frac{1}{h}\left(\frac{2t+2h+1}{t+h+3} - \frac{2t+1}{t+3}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3)}{(t+h+3)(t+3)} - \frac{(2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{(2t+2h+1)(t+3) - (2t+1)(t+h+3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + t + 6t + 6h + 3 - (2t^2 + 2th + 6t +t + h + 3)}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{2t^2 + 2ht + 7t + 6h + 3 - 2t^2 - 2th - 7t -h - 3}{(t+3)(t+h+3)}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{1}{h}\left(\frac{5h}{(t+3)(t+h+3)}\right).
\end{align*}$$
Nothing but algebra so far. 
Can you take it from here?
If you made a mistake in your own derivation, have you spotted it? 
