Answer
$$q'=\csc^2\Big(\frac{\sin t}{t}\Big)\frac{\sin t-t\cos t}{t^2}$$
Work Step by Step
$$q=\cot\Big(\frac{\sin t}{t}\Big)$$
The derivative of function $q$ is $$q'=-\csc^2\Big(\frac{\sin t}{t}\Big)\Big(\frac{\sin t}{t}\Big)'=-\csc^2\Big(\frac{\sin t}{t}\Big)\frac{(\sin t)'t-\sin t(t)'}{t^2}$$
$$q'=-\csc^2\Big(\frac{\sin t}{t}\Big)\frac{t\cos t-\sin t}{t^2}$$
$$q'=\csc^2\Big(\frac{\sin t}{t}\Big)\frac{\sin t-t\cos t}{t^2}$$
