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We are given:

\[
u = 4x + 2y, \quad v = 2x - 4y
\]

We need to find the value of the following product:

\[
\left(\frac{\partial u}{\partial x}\right)_v \cdot \left(\frac{\partial x}{\partial u}\right)_v \cdot \left(\frac{\partial v}{\partial y}\right)_u \cdot \left(\frac{\partial y}{\partial v}\right)_x
\]

### Step 1: Compute partial derivatives

1. **\(\frac{\partial u}{\partial x}\):**
\[
u = 4x + 2y \quad \Rightarrow \quad \frac{\partial u}{\partial x} = 4
\]

2. **\(\frac{\partial v}{\partial y}\):**
\[
v = 2x - 4y \quad \Rightarrow \quad \frac{\partial v}{\partial y} = -4
\]

### Step 2: Express \(\frac{\partial x}{\partial u}\) and \(\frac{\partial y}{\partial v}\) using implicit differentiation.

We need to express \(x\) and \(y\) in terms of \(u\) and \(v\) from the given system of equations:

\[
u = 4x + 2y \quad \text{and} \quad v = 2x - 4y
\]

Solve these equations simultaneously.

From the second equation, express \(x\) in terms of \(v\) and \(y\):
\[
v = 2x - 4y \quad \Rightarrow \quad x = \frac{v + 4y}{2}
\]

Substitute this expression for \(x\) into the first equation:
\[
u = 4\left(\frac{v + 4y}{2}\right) + 2y = 2(v + 4y) + 2y = 2v + 8y + 2y = 2v + 10y
\]
Thus, we have:
\[
u = 2v + 10y \quad \Rightarrow \quad y = \frac{u - 2v}{10}
\]

Now substitute this \(y\) back into the expression for \(x\):
\[
x = \frac{v + 4\left(\frac{u - 2v}{10}\right)}{2} = \frac{v + \frac{4u - 8v}{10}}{2} = \frac{5v + 4u - 8v}{10} = \frac{4u - 3v}{10}
\]

### Step 3: Calculate remaining partial derivatives

Now that we have \(x\) and \(y\) in terms of \(u\) and \(v\), compute the derivatives:

3. **\(\frac{\partial x}{\partial u}\):**
\[
x = \frac{4u - 3v}{10} \quad \Rightarrow \quad \frac{\partial x}{\partial u} = \frac{4}{10} = \frac{2}{5}
\]

4. **\(\frac{\partial y}{\partial v}\):**
\[
y = \frac{u - 2v}{10} \quad \Rightarrow \quad \frac{\partial y}{\partial v} = -\frac{2}{10} = -\frac{1}{5}
\]

### Step 4: Multiply all derivatives

The product of the partial derivatives is:

\[
\left(\frac{\partial u}{\partial x}\right)_v \cdot \left(\frac{\partial x}{\partial u}\right)_v \cdot \left(\frac{\partial v}{\partial y}\right)_u \cdot \left(\frac{\partial y}{\partial v}\right)_x
= 4 \cdot \frac{2}{5} \cdot (-4) \cdot \left(-\frac{1}{5}\right)
\]

Simplify:

\[
= 4 \cdot \frac{2}{5} \cdot (-4) \cdot \left(-\frac{1}{5}\right) = 4 \cdot \frac{2}{5} \cdot 4 \cdot \frac{1}{5} = \frac{4 \cdot 2 \cdot 4 \cdot 1}{5 \cdot 5} = \frac{32}{25}
\]

### Final Answer:

The value of the given expression is \(\frac{32}{25}\).

Would you like further details or have any questions? 

Here are five related questions for you to explore:
1. How is implicit differentiation applied when working with partial derivatives?
2. What is the Jacobian determinant, and how does it relate to these types of problems?
3. How can simultaneous equations be solved algebraically using matrices?
4. Why does the chain rule extend to functions of several variables in partial derivatives?
5. What is the geometric interpretation of partial derivatives?

**Tip:** Always remember to apply the chain rule carefully when dealing with composite functions of multiple variables.

If u = 4x + 2y, v = 2x - 4y, find the value of (∂u/∂x)_v ⋅ (∂x/∂u)_v ⋅ (∂v/∂y)_u ⋅ (∂y/∂v)_x.

Solve the problem involving partial derivatives where u = 4x + 2y and v = 2x - 4y. Learn how to apply the chain rule and implicit differentiation to find the value of the product (∂u/∂x)_v ⋅ (∂x/∂u)_v ⋅ (∂v/∂y)_u ⋅ (∂y/∂v)_x, step-by-step. Suitable for university-level students studying multivariable calculus.

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