Answer
$$q'=\cos\Big(\frac{t}{\sqrt{t+1}}\Big)\frac{t+2}{2(t+1)\sqrt{t+1}}$$
Work Step by Step
$$q=\sin\Big(\frac{t}{\sqrt{t+1}}\Big)$$
The derivative of function $q$ is $$q'=\cos\Big(\frac{t}{\sqrt{t+1}}\Big)\Big(\frac{t}{\sqrt{t+1}}\Big)'$$
We have $$\Big(\frac{t}{\sqrt{t+1}}\Big)'=\frac{(t)'\sqrt{t+1}-t(\sqrt{t+1})'}{t+1}=\frac{\sqrt{t+1}-\frac{t(t+1)'}{2\sqrt{t+1}}}{t+1}$$ $$=\frac{\sqrt{t+1}-\frac{t}{2\sqrt{t+1}}}{t+1}=\frac{\frac{2(t+1)-t}{2\sqrt{t+1}}}{t+1}=\frac{t+2}{2(t+1)\sqrt{t+1}}$$
Therefore, $$q'=\cos\Big(\frac{t}{\sqrt{t+1}}\Big)\frac{t+2}{2(t+1)\sqrt{t+1}}$$
