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Problem 1: Find the derivative of the function using the definition of derivative.

Given function: $A(p) = 2p^3 + 9p$

To find the derivative $A'(p)$, we use the power rule for differentiation:

1. Differentiate $2p^3$: $\frac{d}{dp}(2p^3) = 2 \cdot 3p^{3-1} = 6p^2$

2. Differentiate $9p$: $\frac{d}{dp}(9p) = 9$

So, the derivative $A'(p)$ is: $A'(p) = 6p^2 + 9$

Problem 2: State the domain of the function $A(p)$.

The function $A(p) = 2p^3 + 9p$ is a polynomial, and polynomials are defined for all real numbers. Thus, the domain of $A(p)$ is: $\text{Domain of } A(p): \ (-\infty, \infty)$

Problem 3: State the domain of the derivative $A'(p)$.

Similarly, the derivative $A'(p) = 6p^2 + 9$ is also a polynomial, and like all polynomials, it is defined for all real numbers. Therefore, the domain of $A'(p)$ is: $\text{Domain of } A'(p): \ (-\infty, \infty)$

Summary of Answers:

1. $A'(p) = 6p^2 + 9$
2. Domain of $A(p)$: $(-\infty, \infty)$
3. Domain of $A'(p)$: $(-\infty, \infty)$

Would you like further details or explanations on any part of this problem?

Here are 5 related questions to expand your understanding:

1. What is the definition of a derivative, and how does it apply to polynomial functions?
2. How would you find the second derivative of $A(p)$?
3. What is the geometric interpretation of the derivative $A'(p)$?
4. Could there be any cases where the domain of a derivative differs from the domain of the original function?
5. How does the degree of a polynomial affect the degree of its derivative?

Tip: Remember that the derivative of a polynomial of degree $n$ will be a polynomial of degree $n-1$.