It looks like you need help with two integrals involving derivatives.

Let’s break down both parts:

Part A:

The problem asks you to evaluate:

$\frac{d}{dz} \int_{9}^{z^2} \ln(x^3 + 5) \, dx$

This requires the Fundamental Theorem of Calculus, where we differentiate an integral with variable limits. The formula to use is:

$\frac{d}{dz} \int_{a(z)}^{b(z)} f(x) \, dx = f(b(z)) \cdot b'(z) - f(a(z)) \cdot a'(z)$

Here, the upper limit is $z^2$ and the lower limit is $9$. So, we differentiate the integral with respect to $z$:

1. The derivative of the upper limit $z^2$ is $2z$.
2. The lower limit is a constant $9$, so its derivative is zero.

Now, apply this to your integral:

$\frac{d}{dz} \int_{9}^{z^2} \ln(x^3 + 5) \, dx = \ln\left((z^2)^3 + 5\right) \cdot 2z$

Simplifying:

$= \ln(z^6 + 5) \cdot 2z$

So, the solution for part A is:

$\ln(z^6 + 5) \cdot 2z$

Part B:

This part involves:

$\frac{d}{dv} \int_{14}^{\sin(v)} \arctan(w - 8) \, dw$

Here, the upper limit is $\sin(v)$ and the lower limit is $14$. Using the same method as part A:

1. The derivative of the upper limit $\sin(v)$ is $\cos(v)$.
2. The lower limit $14$ is constant, so its derivative is zero.

Now, apply this:

$\frac{d}{dv} \int_{14}^{\sin(v)} \arctan(w - 8) \, dw = \arctan\left(\sin(v) - 8\right) \cdot \cos(v)$

So, the solution for part B is:

$\arctan\left(\sin(v) - 8\right) \cdot \cos(v)$

Final Answers:

• A: $\ln(z^6 + 5) \cdot 2z$
• B: $\arctan\left(\sin(v) - 8\right) \cdot \cos(v)$

Would you like further clarification on any part of this? Or is there something more specific you'd like to explore?

Here are a few questions that may expand on this topic:

1. What is the Fundamental Theorem of Calculus and how does it apply to derivative-integral problems?
2. Can you differentiate more complex integrals with varying functions inside the integrals?
3. How do we handle trigonometric functions as limits in integrals?
4. What are some real-life applications of the Fundamental Theorem of Calculus?
5. How do you integrate and differentiate inverse trigonometric functions?

Tip: When you encounter integrals with variable limits, always remember the chain rule and apply it carefully to each part of the derivative.