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!