To solve for \( y'(2) \), we need to compute the derivative \( y'(x) \) and then evaluate it at \( x = 2 \).

Given:

\[
y = 5 + \frac{2x^3 + 8x}{(2,2)}
\]

It seems there is a formatting issue. If you clarify what the part "(2,2)" means, that would help. If it's meant to divide the expression, you might be referring to:

\[
y = 5 + \frac{2x^3 + 8x}{2}
\]

In that case, the expression for \( y \) would be:

\[
y = 5 + \frac{1}{2}(2x^3 + 8x)
\]

Let's simplify:

\[
y = 5 + x^3 + 4x
\]

### Step 1: Differentiate \( y \) with respect to \( x \):

\[
y'(x) = \frac{d}{dx} \left( 5 + x^3 + 4x \right) = 3x^2 + 4
\]

### Step 2: Evaluate \( y'(x) \) at \( x = 2 \):

\[
y'(2) = 3(2)^2 + 4 = 3(4) + 4 = 12 + 4 = 16
\]

Thus, \( y'(2) = 16 \).

If you'd like to confirm the original expression or have further clarifications, feel free to ask!

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Would you like more details, or do you have any questions? Here are 5 related questions:

1. How do you calculate higher-order derivatives?
2. Can you explain the power rule for differentiation?
3. What is implicit differentiation, and how does it differ from regular differentiation?
4. How do you differentiate products of functions (product rule)?
5. How do you find the second derivative of a function?

**Tip:** Always simplify your function as much as possible before differentiating to avoid unnecessary complications.

Learn how to find the derivative of a polynomial function and evaluate y'(2) for the expression y = 5 + (2x^3 + 8x)/2. This solution includes key calculus concepts such as the power rule for differentiation. Suitable for students in Grades 11-12.

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