We are given the function $M = xe^y - z^2$, where $x = 2uv$, $y = u - v$, and $z = u + v$, and we need to compute the partial derivatives $\frac{\partial M}{\partial u}$ and $\frac{\partial M}{\partial v}$ using the Chain Rule.

Step 1: Chain Rule Setup

By the chain rule for multivariable functions, we express the total derivative of $M$ with respect to $u$ and $v$ as follows: $\frac{\partial M}{\partial u} = \frac{\partial M}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial M}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial M}{\partial z} \cdot \frac{\partial z}{\partial u}$ $\frac{\partial M}{\partial v} = \frac{\partial M}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial M}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial M}{\partial z} \cdot \frac{\partial z}{\partial v}$

Step 2: Compute Partial Derivatives of $M$

The partial derivatives of $M = xe^y - z^2$ are: $\frac{\partial M}{\partial x} = e^y$ $\frac{\partial M}{\partial y} = xe^y$ $\frac{\partial M}{\partial z} = -2z$

Step 3: Compute Partial Derivatives of $x$, $y$, and $z$

We also compute the partial derivatives of $x = 2uv$, $y = u - v$, and $z = u + v$: $\frac{\partial x}{\partial u} = 2v, \quad \frac{\partial x}{\partial v} = 2u$ $\frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1$ $\frac{\partial z}{\partial u} = 1, \quad \frac{\partial z}{\partial v} = 1$

Step 4: Evaluate at $u = 3$, $v = -1$

We first substitute $u = 3$ and $v = -1$ into the expressions for $x$, $y$, and $z$: $x = 2uv = 2(3)(-1) = -6$ $y = u - v = 3 - (-1) = 4$ $z = u + v = 3 + (-1) = 2$

Now, substitute these values into the partial derivatives of $M$: $\frac{\partial M}{\partial x} = e^y = e^4$ $\frac{\partial M}{\partial y} = xe^y = (-6)e^4$ $\frac{\partial M}{\partial z} = -2z = -2(2) = -4$

Step 5: Compute $\frac{\partial M}{\partial u}$

Substitute everything into the chain rule for $\frac{\partial M}{\partial u}$: $\frac{\partial M}{\partial u} = e^4(2v) + (-6e^4)(1) + (-4)(1)$ Substitute $v = -1$: $\frac{\partial M}{\partial u} = e^4(2(-1)) + (-6e^4) + (-4)$ $\frac{\partial M}{\partial u} = -2e^4 - 6e^4 - 4 = -8e^4 - 4$

Step 6: Compute $\frac{\partial M}{\partial v}$

Now, substitute into the chain rule for $\frac{\partial M}{\partial v}$: $\frac{\partial M}{\partial v} = e^4(2u) + (-6e^4)(-1) + (-4)(1)$ Substitute $u = 3$: $\frac{\partial M}{\partial v} = e^4(2(3)) + (-6e^4)(-1) + (-4)$ $\frac{\partial M}{\partial v} = 6e^4 + 6e^4 - 4 = 12e^4 - 4$

Final Answers

$\frac{\partial M}{\partial u} = -8e^4 - 4$ $\frac{\partial M}{\partial v} = 12e^4 - 4$

Let me know if you'd like more details on any of the steps or if you have any questions!

Related Questions:

1. How does the chain rule apply to multivariable functions?
2. How do partial derivatives differ from total derivatives?
3. What is the role of the exponential function in the chain rule?
4. How does the structure of $x$, $y$, and $z$ affect the final result?
5. What are the practical applications of partial derivatives?

Tip:

When using the chain rule, it's important to carefully track which variables depend on which others, especially in complex expressions involving multiple functions.