To find the derivative of this function, there is a property that we should know called the Constant Multiple Rule, which says:
[tex]\dfrac{d}{dx}[cf(x)] = cf'(x)[/tex] (where [tex]c[/tex] is a constant)
Remember that the derivative of [tex]\csc(x)[/tex] is [tex]-\csc(x)\cot(x)[/tex]. However, you may notice that we are finding the derivative of [tex]\dfrac{1}{2}\csc(2x)[/tex], not [tex]\dfrac{1}{2} \csc(x)[/tex]. So, we are going to have to use the chain rule. To complete the chain rule for the derivative of a trigonometric function (in layman's terms) is basically the following: First, complete the derivative of the trig function as you would if what was inside the trig function is [tex]x[/tex]. Then, take the derivative of what's inside of the trig function and multiply it by what you found in the first step.
Let's apply that to our problem. Right now, I am not going to worry about the [tex]\dfrac{1}{2}[/tex] at the front of the equation, since we can just multiply it back in at the end of our problem. So, let's examine [tex]\csc(2x)[/tex]. We see that what's inside the trig function is [tex]2x[/tex], which has a derivative of 2. Thus, let's first find the derivative of [tex]\csc(2x)[/tex] as if [tex]2x[/tex] was just [tex]x[/tex] and then multiply it by 2.
The derivative of [tex]\csc(2x)[/tex] would first be [tex]-\cot(2x)\csc(2x)[/tex]. Multiplying it by 2, we get our derivative of [tex]-2\cot(2x)\csc(2x)[/tex]. However, don't forget to multiply it by the [tex]\dfrac{1}{2}[/tex] that we removed near the beginning. This gives us our final derivative of [tex]-\cot(2x)\csc(2x)[/tex].
Remember that we now have to find the derivative at the given point. To do this, simply "plug in" the point into the derivative using the x-coordinate. This is shown below:
[tex]-\cot[2(\dfrac{\pi}{4})]\csc[2(\dfrac{\pi}{4})][/tex]
[tex]-\cot(\dfrac{\pi}{2})\csc(\dfrac{\pi}{2})[/tex]
[tex]-(0)(1) = \boxed{0}[/tex]
Our final answer is 0.