Yes, the derivative of \( \sec x \) with respect to \( x \) is indeed \( \sec x \tan x \). Let's go through the reasoning behind it:

### Step-by-Step Derivation:

1. **Start with the function:**
   \[
   f(x) = \sec x = \frac{1}{\cos x}
   \]

2. **Apply the quotient rule** to differentiate \( \sec x \), where:
   \[
   \frac{d}{dx} \left( \frac{1}{\cos x} \right) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x}
   \]
   This simplifies to:
   \[
   \frac{\sin x}{\cos^2 x}
   \]

3. **Recognize the trigonometric identity:**
   \[
   \frac{\sin x}{\cos^2 x} = \sec x \tan x
   \]

Therefore:
\[
\frac{d}{dx} \sec x = \sec x \tan x
\]

Would you like more details or have any questions?

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### Relative Questions:
1. What is the derivative of \( \tan x \)?
2. How do you apply the chain rule to trigonometric functions like \( \sec(2x) \)?
3. Can you prove the derivative of \( \csc x \) using similar steps?
4. How does the product rule apply to a function like \( \sec x \cdot \cos x \)?
5. What is the second derivative of \( \sec x \)?

### Tip:
When dealing with derivatives of trigonometric functions, it's helpful to remember their identities and corresponding derivatives, as they often simplify the process!

Understand how to differentiate sec x using the quotient rule and trigonometric identities. This solution provides a detailed, step-by-step explanation of the derivative d/dx(sec x) = sec x tan x, suitable for high school calculus students.

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