We are tasked with finding the second derivative $y''$ for the function:

$y = 8 \sec x$

and using the product rule, not the chain rule.

Step 1: Express the function as a product

We can rewrite $y = 8 \sec x$ as a product:

$y = 8 \cdot \sec x$

Here, $u(x) = 8$ and $v(x) = \sec x$.

Step 2: First derivative $y'$

To differentiate $y$, we apply the product rule:

$y' = u'(x)v(x) + u(x)v'(x)$

• $u(x) = 8$, so $u'(x) = 0$
• $v(x) = \sec x$, and $v'(x) = \sec x \tan x$

Thus:

$y' = 0 \cdot \sec x + 8 \cdot (\sec x \tan x)$

So,

$y' = 8 \sec x \tan x$

Step 3: Second derivative $y''$

Now, differentiate $y' = 8 \sec x \tan x$ using the product rule again. Let’s set $u(x) = 8 \sec x$ and $v(x) = \tan x$.

• $u'(x) = 8 \cdot (\sec x \tan x)$ because $\frac{d}{dx}(\sec x) = \sec x \tan x$
• $v'(x) = \sec^2 x$ because $\frac{d}{dx}(\tan x) = \sec^2 x$

Using the product rule:

$y'' = u'(x)v(x) + u(x)v'(x)$

Substitute the expressions:

$y'' = (8 \sec x \tan x) \cdot \tan x + (8 \sec x) \cdot \sec^2 x$

Simplifying:

$y'' = 8 \sec x \tan^2 x + 8 \sec^3 x$

Thus, the second derivative is:

$y'' = 8 \sec x (\tan^2 x + \sec^2 x)$

This is the final result for the second derivative of the given function.

Would you like more details on this solution or have any other questions?

Here are 5 related questions to expand your understanding:

1. How does the product rule differ from the chain rule?
2. What is the geometric meaning of the second derivative in this context?
3. How can we verify this result by using the chain rule approach instead?
4. What are the applications of second derivatives in real-world problems?
5. Can we explore the second derivative for other trigonometric functions like $\sin x$ or $\cos x$?

Tip: The product rule is essential when you have two functions multiplied together, while the chain rule is used for composite functions. Always identify the type of function before selecting the rule.