To find the derivative of the function \( f(x) = \sec(x) \tan(x) \), we'll use the **product rule** and the known derivatives of basic trigonometric functions.

The product rule states that if \( f(x) = u(x) \cdot v(x) \), then:

\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Here, we have \( u(x) = \sec(x) \) and \( v(x) = \tan(x) \). Let's differentiate each:

- Derivative of \( u(x) = \sec(x) \): 
\[
u'(x) = \sec(x) \tan(x)
\]
- Derivative of \( v(x) = \tan(x) \):
\[
v'(x) = \sec^2(x)
\]

Now applying the product rule:

\[
f'(x) = \sec(x) \tan(x) \cdot \tan(x) + \sec(x) \cdot \sec^2(x)
\]

Simplify the expression:

\[
f'(x) = \sec(x) \tan^2(x) + \sec^3(x)
\]

So the derivative of \( f(x) = \sec(x) \tan(x) \) is:

\[
f'(x) = \sec(x) \tan^2(x) + \sec^3(x)
\]

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### Related Questions:
1. What are the steps to differentiate \( \sec(x) \cdot \sin(x) \)?
2. How do you find the derivative of the function \( f(x) = \sec(x) + \tan(x) \)?
3. Can you explain the derivative of \( \tan(x) \) and why it results in \( \sec^2(x) \)?
4. How does the product rule work for non-trigonometric functions?
5. What happens if you differentiate \( f(x) = \sec^2(x) \tan(x) \)?

### Tip:
Remember, for trigonometric derivatives, it's always good to recall the basic derivatives of \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and \( \sec(x) \), as they often appear in combination!

This page explains how to differentiate f(x) = sec(x)tan(x) using the product rule. Learn step-by-step how to apply the product rule and basic trigonometric derivatives, including sec(x) and tan(x). Suitable for advanced high school or early college calculus students.

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