The given problem involves using the Fundamental Theorem of Calculus to evaluate and differentiate an integral. Here's the breakdown of the solution:

Problem Statement:

Given $g(x) = \int_{2}^{x} t^4 \, dt$, we are tasked to:

1. Find $\frac{d}{dx} g(x)$.
2. Use the Fundamental Theorem of Calculus, evaluate the integral, and then differentiate it.

Solution:

1. Directly Differentiating $g(x)$:

By the Fundamental Theorem of Calculus (Part 1), the derivative of the integral is given by:

$\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x),$

where $f(t)$ is the integrand.

Here, $f(t) = t^4$, so:

$\frac{d}{dx} g(x) = x^4.$

2. Evaluate the Integral, Then Differentiate:

The integral $g(x)$ can be evaluated explicitly:

$g(x) = \int_{2}^{x} t^4 \, dt.$

To compute this:

$\int t^4 \, dt = \frac{t^5}{5} + C.$

Applying the limits $2$ to $x$:

[ g(x) = \left[\frac{x^5}{5}\right] - \left[\frac{2^5}{5}\right] = \frac{x^5}{5} - \frac{32}{5}. ]

Now, differentiate $g(x)$:

$\frac{d}{dx} g(x) = \frac{d}{dx} \left( \frac{x^5}{5} - \frac{32}{5} \right) = \frac{5x^4}{5} = x^4.$

Final Answer:

$\frac{d}{dx} g(x) = x^4.$

Both methods confirm the result.

Do you want further clarification or additional examples? Here are 5 related questions for practice:

1. What is the derivative of $g(x) = \int_{1}^{x} \sin(t) \, dt$?
2. If $g(x) = \int_{3}^{x} (2t + 1) \, dt$, find $g'(x)$.
3. Evaluate $\int_{0}^{x} e^t \, dt$ and then differentiate it.
4. Explain the difference between Fundamental Theorem of Calculus Part 1 and Part 2.
5. Solve $\frac{d}{dx} \int_{x}^{4} \cos(t) \, dt$.

Tip: Always verify the limits of integration when applying the Fundamental Theorem of Calculus!