For Exercise 8, the problem involving an integral is:

$f(x,y) = \int_y^x \cos(e^t) \, dt.$

To find the partial derivatives $f_x(x, y)$ and $f_y(x, y)$, we can apply the Leibniz Rule for differentiation under the integral sign.

1. Partial derivative with respect to $x$:

Since $x$ is the upper limit of the integral, we differentiate as follows:

$f_x(x, y) = \frac{\partial}{\partial x} \int_y^x \cos(e^t) \, dt = \cos(e^x).$

2. Partial derivative with respect to $y$:

Since $y$ is the lower limit of the integral, we differentiate as follows:

$f_y(x, y) = \frac{\partial}{\partial y} \int_y^x \cos(e^t) \, dt = -\cos(e^y).$

Final Answer:

$f_x(x, y) = \cos(e^x) \quad \text{and} \quad f_y(x, y) = -\cos(e^y).$

Would you like more details on this solution, or do you have any further questions?

Here are 5 related questions to further explore the topic:

1. How do you apply the Leibniz Rule for integrals with variable limits?
2. What is the importance of the Fundamental Theorem of Calculus in finding partial derivatives?
3. How would the answer change if the integral bounds were constants instead?
4. Can this approach be extended to integrals with both limits depending on variables?
5. How does the chain rule apply in the context of partial derivatives with integrals?

Tip: Remember that when dealing with integrals with variable limits, you treat the limits as functions of the variables for differentiation purposes.