We are tasked with finding $f'(x)$ where $f(x) = \int_0^{x^2} t^4 \, dt$. This involves applying the Fundamental Theorem of Calculus and the Chain Rule.

Step-by-Step Solution

1. General Structure Using the Fundamental Theorem of Calculus: The derivative of an integral with an upper limit $g(x)$ is given by: $\frac{d}{dx} \int_a^{g(x)} F(t) \, dt = F(g(x)) \cdot g'(x)$ where $F(t) = t^4$ in this case.

2. Substituting the Limits: Here, $f(x) = \int_0^{x^2} t^4 \, dt$. Using the above formula: $f'(x) = t^4 \big|_{t = x^2} \cdot \frac{d}{dx}(x^2)$

3. Evaluating $t^4 \big|_{t = x^2}$: Substitute $t = x^2$: $t^4 = (x^2)^4 = x^8$

4. Computing $\frac{d}{dx}(x^2)$: The derivative of $x^2$ with respect to $x$ is: $\frac{d}{dx}(x^2) = 2x$

5. Combine the Results: Multiply the result from step 3 by the result from step 4: $f'(x) = x^8 \cdot 2x = 2x^9$

Final Answer:

$f'(x) = 2x^9$

Would you like further explanation or a related example?

5 Related Questions:

1. How does the Fundamental Theorem of Calculus connect derivatives and integrals?
2. What happens if the lower limit of the integral is a function of $x$ instead of the upper limit?
3. How would $f'(x)$ change if the integral bounds were reversed?
4. Can this approach be extended to integrals with both bounds being functions of $x$?
5. What is the physical interpretation of taking the derivative of an integral in applications like physics?

Tip:

When differentiating an integral with variable limits, always carefully handle the Chain Rule for the variable bounds.