To find the derivative of $f(x) = \frac{\csc(x)}{\tan(x)}$, we will use the quotient rule for differentiation.

The quotient rule states that if we have a function of the form:

$f(x) = \frac{g(x)}{h(x)},$

then its derivative is given by:

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$

In this case:

• $g(x) = \csc(x)$
• $h(x) = \tan(x)$

Step 1: Find the derivatives of $g(x)$ and $h(x)$:

1. The derivative of $g(x) = \csc(x)$ is: $g'(x) = -\csc(x) \cot(x)$

2. The derivative of $h(x) = \tan(x)$ is: $h'(x) = \sec^2(x)$

Step 2: Apply the quotient rule:

$f'(x) = \frac{(-\csc(x) \cot(x)) \tan(x) - \csc(x) \sec^2(x)}{[\tan(x)]^2}.$

Step 3: Simplify the expression:

$f'(x) = \frac{-\csc(x) \cot(x) \tan(x) - \csc(x) \sec^2(x)}{\tan^2(x)}.$

Now, let's simplify the terms inside the numerator:

• Recall that $\cot(x) = \frac{1}{\tan(x)}$, so: $\csc(x) \cot(x) \tan(x) = \csc(x).$

Thus, the expression simplifies to:

$f'(x) = \frac{-\csc(x) - \csc(x) \sec^2(x)}{\tan^2(x)}.$

You can factor out $-\csc(x)$ from the numerator:

$f'(x) = \frac{-\csc(x) \left(1 + \sec^2(x)\right)}{\tan^2(x)}.$

This is the derivative of $f(x)$.

Final Answer:

$f'(x) = \frac{-\csc(x) \left(1 + \sec^2(x)\right)}{\tan^2(x)}.$

Let me know if you need any further clarification or additional steps!

Follow-up Questions:

1. Can you explain the quotient rule in more detail with another example?
2. What is the relationship between $\csc(x)$ and $\sec(x)$?
3. How would the derivative change if the function was $f(x) = \frac{\sin(x)}{\tan(x)}$?
4. Can we simplify this derivative further, or is this the most reduced form?
5. How do we find the critical points of this function from the derivative?

Tip:

When simplifying trigonometric expressions in derivatives, it helps to recall the basic identities like $\cot(x) = \frac{1}{\tan(x)}$ and $\sec^2(x) = 1 + \tan^2(x)$, as they can simplify the terms significantly.