We are given $f(x) = \sec x$ and asked to find the second derivative $f''(x)$.

Step 1: Find the first derivative $f'(x)$

The derivative of $\sec x$ is: $f'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x$

Step 2: Find the second derivative $f''(x)$

Now, differentiate $f'(x) = \sec x \tan x$ using the product rule: $f''(x) = \frac{d}{dx}(\sec x \tan x)$ Applying the product rule: $f''(x) = \frac{d}{dx}(\sec x) \cdot \tan x + \sec x \cdot \frac{d}{dx}(\tan x)$

From earlier, we know: $\frac{d}{dx}(\sec x) = \sec x \tan x$ and the derivative of $\tan x$ is $\sec^2 x$.

Thus: $f''(x) = (\sec x \tan x) \cdot \tan x + \sec x \cdot \sec^2 x$

Simplifying: $f''(x) = \sec x \tan^2 x + \sec^3 x$

So, the second derivative is: $f''(x) = \sec x (\tan^2 x + \sec^2 x)$

Final Answer:

$f''(x) = \sec x (\sec^2 x)$ $f''(x) = \sec^3 x$

Would you like any further details or explanations?

Here are 5 related questions for deeper understanding:

1. What is the third derivative of $f(x) = \sec x$?
2. How can you apply the chain rule to find derivatives of trigonometric functions?
3. What is the integral of $\sec x$?
4. How can we find higher-order derivatives for other trigonometric functions like $\sin x$ or $\cos x$?
5. Can you explain the product rule and how it was used in this problem?

Tip: Always remember to apply the product rule when differentiating the product of two functions.