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The image shows an exercise from a textbook related to using the chain rule in partial differentiation to find the total derivative \( \frac{dz}{dt} \) at a given point.

### Problem Breakdown:
1. **Part (a)**: 
   \[
   z = x \cos(y), \quad \text{where} \quad x = \sin(t), \quad y = \cos(t)
   \]
   Find \( \frac{dz}{dt} \).

2. **Part (b)**:
   \[
   z = x^2 + 2xy + y^2, \quad \text{where} \quad x = e^t \cos(t), \quad y = e^t \sin(t)
   \]
   Find \( \frac{dz}{dt} \).

### Solutions:

#### Part (a)
1. **Expression for \( z \)**:
   \[
   z = x \cos(y)
   \]
2. **Given functions**:
   \[
   x = \sin(t), \quad y = \cos(t)
   \]

   The total derivative \( \frac{dz}{dt} \) using the chain rule is given by:
   \[
   \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}
   \]
   
3. **Partial derivatives**:
   \[
   \frac{\partial z}{\partial x} = \cos(y)
   \]
   \[
   \frac{\partial z}{\partial y} = -x \sin(y)
   \]
   
4. **Derivatives of \( x \) and \( y \) with respect to \( t \)**:
   \[
   \frac{dx}{dt} = \cos(t)
   \]
   \[
   \frac{dy}{dt} = -\sin(t)
   \]

5. **Substitute and simplify**:
   \[
   \frac{dz}{dt} = \cos(y) \cdot \cos(t) + \left(-x \sin(y)\right) \cdot (-\sin(t))
   \]
   \[
   \frac{dz}{dt} = \cos(\cos(t)) \cdot \cos(t) + \sin(t) \cdot \sin(\cos(t)) \cdot \sin(t)
   \]
   \[
   \frac{dz}{dt} = \cos(\cos(t)) \cdot \cos(t) + \sin^2(t) \cdot \sin(\cos(t))
   \]

#### Part (b)
1. **Expression for \( z \)**:
   \[
   z = x^2 + 2xy + y^2
   \]
2. **Given functions**:
   \[
   x = e^t \cos(t), \quad y = e^t \sin(t)
   \]

   Again, use the chain rule:
   \[
   \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}
   \]

3. **Partial derivatives**:
   \[
   \frac{\partial z}{\partial x} = 2x + 2y
   \]
   \[
   \frac{\partial z}{\partial y} = 2x + 2y
   \]

4. **Derivatives of \( x \) and \( y \) with respect to \( t \)**:
   \[
   \frac{dx}{dt} = e^t \cos(t) - e^t \sin(t)
   \]
   \[
   \frac{dy}{dt} = e^t \sin(t) + e^t \cos(t)
   \]

5. **Substitute and simplify**:
   \[
   \frac{dz}{dt} = (2x + 2y) \cdot \frac{dx}{dt} + (2x + 2y) \cdot \frac{dy}{dt}
   \]
   Simplify further by substituting the values of \( x \) and \( y \) and their derivatives.

Would you like further steps for the simplification, or do you have any specific questions?

---

**Follow-up Questions:**

1. How does the chain rule apply when dealing with multiple variables?
2. Can you derive the partial derivatives of a more complex function?
3. What are the key differences between total and partial derivatives?
4. How would you handle a similar problem with more variables?
5. What happens if one of the functions is implicitly defined?

**Tip:** Always keep track of the variables being differentiated to avoid mistakes, especially when multiple variables depend on the same parameter.

Explore how to solve partial differentiation problems using the chain rule in calculus. This detailed solution covers finding total derivatives at specific points, including step-by-step calculations for advanced college-level students.

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