Answer
$$G'(t)=\frac{-7}{(3+t)^2}$$
The domain of both $G(t)$ and $G'(t)$ is $(-\infty,-3)U(-3,\infty)$
Work Step by Step
$$G(t)=\frac{1-2t}{3+t}$$
According to definition, $$G'(t)=\lim\limits_{h\to0}\frac{G(t+h)-G(t)}{h}$$
$$G'(t)=\lim\limits_{h\to0}\frac{\frac{1-2(t+h)}{3+t+h}-\frac{1-2t}{3+t}}{h}$$
$$G'(t)=\lim\limits_{h\to0}\frac{[1-2(t+h)](3+t)-(1-2t)(3+t+h)}{h(3+t+h)(3+t)}$$
$$G'(t)=\lim\limits_{h\to0}\frac{(1-2t-2h)(3+t)-(3+t+h-6t-2t^2-2th)}{h(3+t+h)(3+t)}$$
$$G'(t)=\lim\limits_{h\to0}\frac{3+t-6t-2t^2-6h-2th-3-t-h+6t+2t^2+2th}{h(3+t+h)(3+t)}$$
$$G'(t)=\lim\limits_{h\to0}\frac{-7h}{h(3+t+h)(3+t)}$$
$$G'(t)=\lim\limits_{h\to0}\frac{-7}{(3+t+h)(3+t)}$$
$$G'(t)=\frac{-7}{(3+t+0)(3+t)}$$
$$G'(t)=\frac{-7}{(3+t)^2}$$
*Domain of $G(t)$
$G(t)$ is defined for $R$ except where $3+t=0$, or $t=-3$.
Therefore, the domain of $G(t)$ is $(-\infty,-3)U(-3,\infty)$
*Domain of $G'(t)$
$G'(t)$ is defined for $R$ except where $(3+t)^2=0$, or $t=-3$.
Therefore, the domain of $G'(t)$ is $(-\infty,-3)U(-3,\infty)$
