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To find the derivative of \( y = \cos(x) \tan(x) \), we will apply the product rule, which is used to differentiate the product of two functions. The product rule is:

\[
\frac{d}{dx} \left[ u(x) v(x) \right] = u'(x) v(x) + u(x) v'(x)
\]

Here, let:
- \( u(x) = \cos(x) \) (whose derivative is \( u'(x) = -\sin(x) \)),
- \( v(x) = \tan(x) \) (whose derivative is \( v'(x) = \sec^2(x) \)).

Now, applying the product rule:

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

Thus, the derivative is:

\[
\frac{dy}{dx} = \cos(x) \sec^2(x) - \sin(x) \tan(x)
\]

This corresponds to the fourth option in the list.

Would you like further details or have any other questions? Here are five related questions:

1. What is the product rule and when should it be applied?
2. How do you differentiate \( \tan(x) \)?
3. What is the derivative of \( \sin(x) \cdot \cos(x) \)?
4. Can the quotient rule be used in this problem instead of the product rule? Why or why not?
5. How would you integrate \( \cos(x) \cdot \tan(x) \)?

**Tip**: Always remember the basic trigonometric derivatives, such as \( \frac{d}{dx} \tan(x) = \sec^2(x) \), for quick differentiation of trigonometric functions.

Learn how to differentiate the function y = cos(x)tan(x) using the product rule. This problem covers key differentiation concepts, including trigonometric function derivatives. Suitable for students in Grades 10-12.

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