Find the indicated derivative Find the indicated derivative
$\frac {d}{dt}$ $\frac {(6t-5)^6}{t+9}$ 
I'm stuck after getting to this part
$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}$
How do they get to the answer
$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$

They simply factor out $(6t-5)^5$:
$$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}=\frac{(6t−5)^5(6t−5-36(t+9))}{(t+9)^2}.
$$
(Note there seems o be a sign error in your computation)
Faster with logarithmic differentiation:
$$\frac{f'(x)}{f(x)}=\frac{36}{6t-5}-\frac1{t+9}=\frac{30t+329}{(6t-5)(t+9)}, $$
so
$$f'(x)=\frac{30t+329}{(6t-5)(t+9)}\cdot f(x)=\frac{(6t-5)^5(30t+329)}{(t+9)^2}.$$
