If
f ( x )

The given function is:

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

To find $f'(x)$, we need to apply the Leibniz rule for differentiating an integral with variable limits of integration. This rule is given as:

$\frac{d}{dx} \left( \int_{a(x)}^{b(x)} g(t) \, dt \right) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$

In this case:

• $g(t) = t^4$
• $a(x) = x^2$, so $a'(x) = 2x$
• $b(x) = x$, so $b'(x) = 1$

Using the Leibniz rule:

$f'(x) = t^4 \bigg|_{t=x} \cdot (1) - t^4 \bigg|_{t=x^2} \cdot (2x)$

This simplifies to:

$f'(x) = x^4 - (x^2)^4 \cdot 2x$

$f'(x) = x^4 - 2x \cdot x^8$

$f'(x) = x^4 - 2x^9$

Thus, the derivative is:

$f'(x) = x^4 - 2x^9$

Do you want more details or have any questions?

Here are 5 related questions for further practice:

1. What is the derivative of $\int_0^x t^4 \, dt$?
2. Can you apply the Leibniz rule for an integral where both limits depend on $x$?
3. What happens if $f(x) = \int_1^{x^3} \sin(t) \, dt$? What is $f'(x)$?
4. How would you compute $f'(x)$ if the integrand was $t^n$?
5. What is the integral of $t^4$ with respect to $t$?

Tip: When differentiating an integral with variable limits, always apply the Leibniz rule carefully to account for how the limits depend on $x$.