To find the derivative of the function \( f(x) = \frac{\sin(x)}{\cos(2x)} \), we will use the **quotient rule**. The quotient rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their quotient is given by:

\[
\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}
\]

In your case:
- \( u(x) = \sin(x) \)
- \( v(x) = \cos(2x) \)

### Step 1: Find \( u'(x) \) and \( v'(x) \)

- \( u'(x) = \frac{d}{dx}[\sin(x)] = \cos(x) \)
- \( v'(x) = \frac{d}{dx}[\cos(2x)] \)

To differentiate \( \cos(2x) \), we apply the chain rule:
\[
\frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot 2 = -2\sin(2x)
\]

### Step 2: Apply the quotient rule

Now, apply the quotient rule formula:

\[
\frac{d}{dx}\left(\frac{\sin(x)}{\cos(2x)}\right) = \frac{\cos(2x) \cdot \cos(x) - \sin(x) \cdot (-2\sin(2x))}{[\cos(2x)]^2}
\]

Simplify the expression:

\[
= \frac{\cos(2x) \cdot \cos(x) + 2\sin(x) \sin(2x)}{[\cos(2x)]^2}
\]

### Final Answer:

\[
\frac{d}{dx}\left(\frac{\sin(x)}{\cos(2x)}\right) = \frac{\cos(2x) \cdot \cos(x) + 2\sin(x) \sin(2x)}{[\cos(2x)]^2}
\]

Let me know if you need further clarification or have any questions!

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Here are five related questions:
1. How do you differentiate trigonometric functions using the chain rule?
2. Can you explain the quotient rule in detail with an example?
3. How does the chain rule apply to \( \cos(2x) \)?
4. What is the derivative of \( \cos(x) \) with respect to \( x \)?
5. How would you simplify the resulting derivative if there were common terms?

Tip: When applying the chain rule or quotient rule, always be sure to check that you correctly differentiate each component before plugging it into the formula!

Learn how to find the derivative of sin(x) over cos(2x) using the quotient rule and the chain rule. This detailed solution covers key concepts in differentiation, including trigonometric derivatives and applying the quotient rule. Suitable for students in Grades 11-12.

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