## Thomas' Calculus 13th Edition

$\frac{-2}{(2t+1)^2}$
Step 1. Given $f(t)=\frac{1}{2t+1}$, we need to find the limit $f'(t)=\lim_{h\to0}\frac{f(t+h)-f(t)}{h}$ Step 2. We can find $f(t+h)=\frac{1}{2t+2h+1}$ and $f(t+h)-f(t)=\frac{1}{2t+2h+1}-\frac{1}{2t+1}=\frac{2t+1-2t-2h-1}{(2t+2h+1)(2t+1)}=\frac{-2h}{(2t+2h+1)(2t+1)}$ Step 3. Thus $f'(t)=\lim_{h\to0}\frac{f(t+h)-f(t)}{h}=\lim_{h\to0}\frac{-2}{(2t+2h+1)(2t+1)}=\frac{-2}{(2t+1)^2}$