The image contains a problem where the function $f(x) = \csc(x)$ is given, and you are asked to:

(a) Find the derivative of $f(x) = \csc(x)$.

The derivative of the cosecant function is:

$f'(x) = -\csc(x) \cdot \cot(x)$

(b) Evaluate the derivative at $x = \frac{11\pi}{6}$.

To find $f'\left(\frac{11\pi}{6}\right)$, we substitute $x = \frac{11\pi}{6}$ into the derivative formula:

$f'\left(\frac{11\pi}{6}\right) = -\csc\left(\frac{11\pi}{6}\right) \cdot \cot\left(\frac{11\pi}{6}\right)$

Now, we calculate:

• $\csc\left(\frac{11\pi}{6}\right) = \frac{1}{\sin\left(\frac{11\pi}{6}\right)} = \frac{1}{-\frac{1}{2}} = -2$
• $\cot\left(\frac{11\pi}{6}\right) = \frac{\cos\left(\frac{11\pi}{6}\right)}{\sin\left(\frac{11\pi}{6}\right)} = \frac{\sqrt{3}/2}{-\frac{1}{2}} = -\sqrt{3}$

Thus,

$f'\left(\frac{11\pi}{6}\right) = -(-2)(-\sqrt{3}) = -2\sqrt{3}$

So, the value of the derivative at $x = \frac{11\pi}{6}$ is:

$f'\left(\frac{11\pi}{6}\right) = -2\sqrt{3}$

Would you like further clarification or details on this?

Related Questions:

1. What is the derivative of $\csc(x)$ using first principles?
2. How is the derivative of $\cot(x)$ related to $\csc(x)$?
3. Can you find the second derivative of $\csc(x)$?
4. What is the general process to find the derivative of trigonometric functions?
5. How would the derivative change if the function was $\csc(2x)$?

Tip:

Always double-check trigonometric values on the unit circle for angles like $\frac{11\pi}{6}$ to avoid errors when evaluating derivatives.