Answer
$q'=\frac{t+2}{2(t+1)\sqrt{t+1}}\cos\left(\frac{t}{\sqrt{t+1}}\right)$
Work Step by Step
To find the derivative of the function $q=\sin\left(\frac{t}{\sqrt{t+1}}\right)$, we can use the chain rule: $$\begin{aligned}
q'& = \frac{d}{dt} \left(\frac{t}{\sqrt{t+1}}\right)\cos\left(\frac{t}{\sqrt{t+1}}\right)\\
&=\frac{\sqrt{t+1}-\frac{t}{2\sqrt{t+1}}}{t+1}\cos\left(\frac{t}{\sqrt{t+1}}\right)\\
&=\frac{2(t+1)-t}{2(t+1)\sqrt{t+1}}\cos\left(\frac{t}{\sqrt{t+1}}\right)\\
&=\frac{t+2}{2(t+1)\sqrt{t+1}}\cos\left(\frac{t}{\sqrt{t+1}}\right).
\end{aligned}$$ Therefore, the derivative of $q$ is $$q' =\frac{t+2}{2(t+1)\sqrt{t+1}}\cos\left(\frac{t}{\sqrt{t+1}}\right).$$
