To find the derivative $g'(x)$ of the function $g(x) = \frac{\cos(x)}{\sin(x)}$, we can use the quotient rule. Recall that the quotient rule is:

$\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

For the function $g(x) = \frac{\cos(x)}{\sin(x)}$, let:

• $u(x) = \cos(x)$
• $v(x) = \sin(x)$

Now, compute the derivatives:

• $u'(x) = -\sin(x)$
• $v'(x) = \cos(x)$

Applying the quotient rule:

$g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} = \frac{(-\sin(x))\sin(x) - \cos(x)\cos(x)}{\sin^2(x)}$

Simplify the expression in the numerator:

$g'(x) = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}$

Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$:

$g'(x) = \frac{-(\sin^2(x) + \cos^2(x))}{\sin^2(x)} = \frac{-1}{\sin^2(x)}$

So the correct derivative is:

$g'(x) = -\frac{1}{\sin^2(x)}$

This corresponds to Choice C.

Would you like any further explanations?

Here are 5 related questions to explore this concept further:

1. How would you find the second derivative $g''(x)$ of this function?
2. What is the derivative of $g(x) = \cot(x)$, and how does it relate to this problem?
3. Can you apply the chain rule to simplify $g'(x)$ in a different form?
4. How does the derivative $g'(x)$ change at specific values, such as $x = \frac{\pi}{4}$?
5. What happens to $g(x)$ and $g'(x)$ as $x$ approaches 0 or $\pi$?

Tip: Remember that trigonometric identities, like the Pythagorean identity, can simplify your work when dealing with derivatives involving trigonometric functions.