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The question asks to prove that the derivative of \( f(x) = \sec x \) is \( f'(x) = \sec x \tan x \).

### Solution:

Given:
\[
f(x) = \sec x
\]

To find \( f'(x) \), we can use the derivative of \( \sec x \) with respect to \( x \).

#### Derivative of \( \sec x \):
We know:
\[
\frac{d}{dx} (\sec x) = \sec x \tan x
\]

#### Proof:

1. Rewrite \( \sec x \) as \( \frac{1}{\cos x} \).
   
   So,
   \[
   f(x) = \sec x = \frac{1}{\cos x}
   \]

2. Use the quotient rule for differentiation: If \( h(x) = \frac{u(x)}{v(x)} \), then
   \[
   h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
   \]

   Here, \( u(x) = 1 \) and \( v(x) = \cos x \).

3. Compute \( u'(x) \) and \( v'(x) \):
   \[
   u'(x) = 0
   \]
   \[
   v'(x) = -\sin x
   \]

4. Substitute into the quotient rule:
   \[
   f'(x) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{(\cos x)^2}
   \]
   \[
   f'(x) = \frac{\sin x}{\cos^2 x}
   \]

5. Rewrite \( \frac{\sin x}{\cos^2 x} \) using trigonometric identities:
   \[
   f'(x) = \sec x \tan x
   \]

Thus, we have shown that
\[
f'(x) = \sec x \tan x
\]

Would you like more details or have any questions?

### Related Questions:
1. What is the derivative of \( \tan x \)?
2. How do you prove the derivative of \( \csc x \)?
3. What is the derivative of \( \cot x \)?
4. How is the chain rule applied in trigonometric differentiation?
5. What is the second derivative of \( f(x) = \sec x \)?

### Tip:
Memorizing basic trigonometric derivatives, like those of \( \sin x \), \( \cos x \), \( \tan x \), and \( \sec x \), will make solving these kinds of problems faster.

Buktikan turunan dari f(x) = sec x adalah f'(x) = sec x tan x.

Explore the step-by-step solution to prove that the derivative of f(x) = sec x is f'(x) = sec x tan x. This calculus problem utilizes differentiation rules and trigonometric identities, suitable for Grades 11-12.

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