The steps for solving this problem rely on applying the Fundamental Theorem of Calculus for functions with variable limits of integration. Here’s a breakdown of the solution process:

1. Identify the integral with variable limits: We have an integral $\int_y^x \cos(e^t) \, dt$ where $x$ is the upper limit and $y$ is the lower limit.

2. Differentiate with respect to $x$: To find $\frac{\partial}{\partial x} \int_y^x \cos(e^t) \, dt$, we use the Fundamental Theorem of Calculus. When differentiating an integral with a variable upper limit, $x$, we substitute $t = x$ into the integrand: $\frac{\partial}{\partial x} \int_y^x \cos(e^t) \, dt = \cos(e^x).$

3. Differentiate with respect to $y$: For $\frac{\partial}{\partial y} \int_y^x \cos(e^t) \, dt$, we also use the Fundamental Theorem of Calculus. For the lower limit, we add a negative sign and substitute $t = y$: $\frac{\partial}{\partial y} \int_y^x \cos(e^t) \, dt = -\cos(e^y).$

This approach leverages the properties of definite integrals with variable limits. Would you like more details or have questions about any specific part?

Here are some related questions to explore further:

1. How does the Fundamental Theorem of Calculus apply to integrals with both upper and lower variable limits?
2. Can you generalize this process to functions other than $\cos(e^t)$?
3. What happens if both $x$ and $y$ are functions of another variable?
4. How would the answer change if the integrand had a different exponential function, such as $\cos(t^2)$?
5. What if we took the partial derivative with respect to another variable that both $x$ and $y$ depend on?

Tip: When dealing with integrals with variable limits, always consider which limit you're differentiating with respect to and apply the Fundamental Theorem of Calculus accordingly.