1. Let's start with the problem: We need to find the derivative of the function $f(x) = x^3$ and then calculate the value of this derivative at $x = 8$. 2. The derivative of a function $f(x)$, denoted as $f'(x)$, tells us the rate at which $f(x)$ changes as $x$ changes. 3. For power functions like $x^n$, the rule is: the derivative of $x^n$ is $nx^{n-1}$. 4. Applying this rule to $x^3$, we get: $$f'(x) = 3x^{3-1} = 3x^2$$ 5. Now, to find the derivative value at $x = 8$, substitute $8$ into $f'(x)$: $$f'(8) = 3 imes 8^2 = 3 imes 64 = 192$$ 6. So, the derivative of $x^3$ is $3x^2$, and its value at $x = 8$ is $192$. This means at $x=8$, the function $x^3$ changes at a rate of 192 units per increase in $x$.