The problem asks to calculate the derivative:

$\frac{d}{dx} \left( \int_{x^4}^{e^{11x}} \ln(t) \, dt \right)$

using Part 2 of the Fundamental Theorem of Calculus (FTC).

Part 2 of the Fundamental Theorem of Calculus:

It states that if $F(x) = \int_{a(x)}^{b(x)} f(t) \, dt$, then

$\frac{d}{dx} F(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

where $f(t)$ is the integrand, $a(x)$ and $b(x)$ are the lower and upper limits of integration respectively, and $f'(x)$ is the derivative of the limits with respect to $x$.

Step-by-Step Solution:

Here, $f(t) = \ln(t)$, $a(x) = x^4$, and $b(x) = e^{11x}$.

So, applying the Fundamental Theorem:

$\frac{d}{dx} \left( \int_{x^4}^{e^{11x}} \ln(t) \, dt \right) = \ln(e^{11x}) \cdot \frac{d}{dx}(e^{11x}) - \ln(x^4) \cdot \frac{d}{dx}(x^4)$

1. For the upper limit $b(x) = e^{11x}$: \ln(e^{11x}) = 11x \quad \text{(since \ln(e^u) = u)} and $\frac{d}{dx}(e^{11x}) = 11e^{11x}.$

2. For the lower limit $a(x) = x^4$: $\ln(x^4) = 4 \ln(x)$ and $\frac{d}{dx}(x^4) = 4x^3.$

Final Result:

Now substitute these into the formula:

$\frac{d}{dx} \left( \int_{x^4}^{e^{11x}} \ln(t) \, dt \right) = 11x \cdot 11e^{11x} - 4 \ln(x) \cdot 4x^3$

Simplifying:

$= 121x e^{11x} - 16x^3 \ln(x)$

Thus, the derivative is:

$\boxed{121x e^{11x} - 16x^3 \ln(x)}$

Would you like any further details on the solution?

Related Questions:

1. How does the Fundamental Theorem of Calculus help in differentiating integrals?
2. What happens if one of the limits of integration is a constant rather than a function of $x$?
3. How would the result change if the integrand were a different function, such as $\frac{1}{t}$?
4. Can this method be applied if the integrand involves multiple variables, such as $\ln(t + x)$?
5. How would we approach the problem if the limits were swapped, i.e., $\int_{e^{11x}}^{x^4} \ln(t) dt$?

Tip:

Whenever you differentiate an integral with variable limits, apply the Fundamental Theorem of Calculus carefully, remembering to differentiate both the upper and lower limits separately.