#### Answer

If $g$ is differentiable at $t=7$ then $3g$ is also differentiable at $t=7$.

#### Work Step by Step

$g(t)$ is differentiable at $t=7$ means that $$\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$$ has a defined value.
Now about $3g$, the derivative of $3g$, $(3g)'$, at $t=7$ has the following formula:
$$(3g)'=\lim_{h\to0}\frac{(3g)(h+7)-(3g)(7)}{h}=\lim_{h\to0}\frac{3\Big(g(h+7)\Big)-3(g(7))}{h}$$
$$(3g)'=\lim_{h\to0}3\Big(\frac{g(h+7)-g(7)}{h}\Big)=3\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$$
Since $\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$, as said above, has a defined value, $(3g)'=3\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$ also has a defined value.
This means $3g$ is also differentiable at $t=7$.