Question

(problem 83)

Answer 1

To find the derivative of this function, there is a property that we should know called the Constant Multiple Rule, which says:

$\dfrac{d}{dx}[cf(x)] = cf'(x)$ (where $c$ is a constant)

Remember that the derivative of $\csc(x)$ is $-\csc(x)\cot(x)$. However, you may notice that we are finding the derivative of $\dfrac{1}{2}\csc(2x)$, not $\dfrac{1}{2} \csc(x)$. 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 $x$. 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 $\dfrac{1}{2}$ at the front of the equation, since we can just multiply it back in at the end of our problem. So, let's examine $\csc(2x)$. We see that what's inside the trig function is $2x$, which has a derivative of 2. Thus, let's first find the derivative of $\csc(2x)$ as if $2x$ was just $x$ and then multiply it by 2.

The derivative of $\csc(2x)$ would first be $-\cot(2x)\csc(2x)$. Multiplying it by 2, we get our derivative of $-2\cot(2x)\csc(2x)$. However, don't forget to multiply it by the $\dfrac{1}{2}$ that we removed near the beginning. This gives us our final derivative of $-\cot(2x)\csc(2x)$.

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:

$-\cot[2(\dfrac{\pi}{4})]\csc[2(\dfrac{\pi}{4})]$

$-\cot(\dfrac{\pi}{2})\csc(\dfrac{\pi}{2})$

$-(0)(1) = \boxed{0}$

Our final answer is 0.