Derivative of csc(x) evaluated at a point

Using the standard trigonometric derivative rule, $q'(x) = \frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)$. This is one of the six memorized trig derivatives, and it's important to keep the negative sign, which is a common error point.

Now substitute $x = \pi/2$: since $\csc(\pi/2) = \frac{1}{\sin(\pi/2)} = \frac{1}{1} = 1$ and $\cot(\pi/2) = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0$, we get $q'(\pi/2) = -(1)(0) = 0$.