The problem asks for $f'(x)$ and $f'(3)$ based on the integral definition of $f(x)$, given by:

$f(x) = \int_5^x t^8 \, dt$

Step 1: Use the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if we have:

$f(x) = \int_a^x g(t) \, dt$

then the derivative $f'(x)$ is simply:

$f'(x) = g(x)$

In this case, $g(t) = t^8$, so:

$f'(x) = x^8$

Step 2: Calculate $f'(3)$

Now, we substitute $x = 3$ into the expression for $f'(x)$:

$f'(3) = 3^8 = 6561$

Final Answers:

• $f'(x) = x^8$
• $f'(3) = 6561$

Would you like further details or have any questions?

Related Questions:

1. How does the Fundamental Theorem of Calculus apply to definite integrals?
2. What are the conditions under which you can use the Fundamental Theorem of Calculus?
3. How would $f'(x)$ change if the exponent on $t$ were different (e.g., $t^5$)?
4. How do we compute the second derivative, $f''(x)$, in this case?
5. What happens if we change the lower limit of the integral from 5 to some other value?

Tip:

Always remember that the derivative of an integral with a variable upper limit is the integrand evaluated at that upper limit, according to the Fundamental Theorem of Calculus.