Question

If \(z=f(x,y)\), \(x=e^u+e^{-v}\), and \(y=e^{-u}-e^v\), then:

• A. $\frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$
• B. $\frac{\partial z}{\partial u} + \frac{\partial z}{\partial v} = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$
• C. $\left(\frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}\right)^2 = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$
• D. $\left(\frac{\partial z}{\partial u} - \frac{\partial z}{\partial v}\right)^2 = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$

Hint:

To solve chain-rule equations quickly, write out the derivatives of intermediate variables first. Grouping terms systematically saves calculation time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires using the Chain Rule for partial derivatives of composite functions.
Given that $z$ is a function of $x$ and $y$, and $x, y$ are functions of independent variables $u$ and $v$, we calculate $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}$.
Key Formula or Approach:
The chain rule equations are: \[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \] \[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \]

Step 2: Detailed Explanation:
Given relations: \[ x = e^u + e^{-v} \implies \frac{\partial x}{\partial u} = e^u \quad \text{and} \quad \frac{\partial x}{\partial v} = -e^{-v} \] \[ y = e^{-u} - e^v \implies \frac{\partial y}{\partial u} = -e^{-u} \quad \text{and} \quad \frac{\partial y}{\partial v} = -e^v \] Substitute these partial derivatives into the chain rule formulas: \[ \frac{\partial z}{\partial u} = e^u \frac{\partial z}{\partial x} - e^{-u} \frac{\partial z}{\partial y} \] \[ \frac{\partial z}{\partial v} = -e^{-v} \frac{\partial z}{\partial x} - e^v \frac{\partial z}{\partial y} \] Let us compute the expression $\frac{\partial z}{\partial u} - \frac{\partial z}{\partial v}$: \[ \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} = \left( e^u \frac{\partial z}{\partial x} - e^{-u} \frac{\partial z}{\partial y} \right) - \left( -e^{-v} \frac{\partial z}{\partial x} - e^v \frac{\partial z}{\partial y} \right) \] Rearranging the terms by grouping the coefficients of $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$: \[ = \left( e^u + e^{-v} \right) \frac{\partial z}{\partial x} + \left( e^v - e^{-u} \right) \frac{\partial z}{\partial y} \] Now substitute the original variables $x$ and $y$ back into this expression.
We know: \[ e^u + e^{-v} = x \] \[ e^v - e^{-u} = -(e^{-u} - e^v) = -y \] Substitute these values back: \[ \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} \]

Step 3: Final Answer:
The relation is $\frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} = x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$, which corresponds to Option (A).