To solve the problem:

We are given the function:

$g(t) = \int_0^t \sqrt{x + x^4} \, dx$

We need to find $g'(t)$. According to the Fundamental Theorem of Calculus, if:

$G(t) = \int_a^t f(x) \, dx,$

then:

$G'(t) = f(t).$

For our problem, $f(x) = \sqrt{x + x^4}$. Thus, $g'(t) = \sqrt{t + t^4}$.

Correct answer:

$g'(t) = \sqrt{t + t^4}.$

Let me know if you'd like further explanation!

Here are 5 questions that expand on this topic:

1. What is the Fundamental Theorem of Calculus, and how does it apply to definite integrals?
2. How would $g'(t)$ change if the limits of integration were from $t$ to a constant?
3. What happens if the integrand $\sqrt{x + x^4}$ changes to include a trigonometric function, e.g., $\sqrt{x + x^4 + \sin x}$?
4. Can we directly compute $g(t)$ without differentiating it?
5. How do you handle problems where the upper limit is itself a function of $t$?

Tip:

Always use the Fundamental Theorem of Calculus carefully, ensuring you evaluate the derivative correctly in terms of the variable in the integrand and limits.