To find the instantaneous rate of change of the function $f(x) = x^3 + 1$ at $x = 2$ using the first principle (also known as the definition of the derivative), we use the following formula:

$f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}$

Step 1: Substitute the function into the formula

Given $f(x) = x^3 + 1$, we substitute this into the formula for $f'(x)$:

$f'(x) = \lim_{{h \to 0}} \frac{(x+h)^3 + 1 - (x^3 + 1)}{h}$

Step 2: Simplify the expression

First, expand $(x+h)^3$:

$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$

Now substitute this back into the expression:

$f'(x) = \lim_{{h \to 0}} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 1 - x^3 - 1}{h}$

Simplify the numerator:

$f'(x) = \lim_{{h \to 0}} \frac{3x^2h + 3xh^2 + h^3}{h}$

Factor out $h$ from the numerator:

$f'(x) = \lim_{{h \to 0}} \frac{h(3x^2 + 3xh + h^2)}{h}$

Cancel out the $h$ in the numerator and denominator:

$f'(x) = \lim_{{h \to 0}} (3x^2 + 3xh + h^2)$

Step 3: Evaluate the limit as $h$ approaches 0

$f'(x) = 3x^2 + 3x(0) + 0^2 = 3x^2$

Step 4: Find the derivative at $x = 2$

Now, substitute $x = 2$ into the derivative:

$f'(2) = 3(2)^2 = 3 \times 4 = 12$

So, the instantaneous rate of change of $f(x) = x^3 + 1$ at $x = 2$ is $12$.

Would you like more details or have any questions?

Here are five related questions to explore further:

1. How does the instantaneous rate of change compare with the average rate of change between two points?
2. How would you find the derivative of a more complex function using the first principle?
3. What is the geometric interpretation of the derivative at a specific point on the graph?
4. How does the first principle approach differ from using standard derivative rules?
5. Can you apply the first principle to find the derivative of $f(x) = \sin(x)$ at a specific point?

Tip: The first principle method is foundational for understanding derivatives and can be applied to any differentiable function, making it essential to grasp before moving on to more advanced derivative rules.