Question

If \(z = x^2 - y^2\) then \( \frac{1}{x} \frac{\partial z}{\partial x} + \frac{1}{y} \frac{\partial z}{\partial y} = \)

• A. 1
• B. 2x + 2y
• C. 0
• D. 2x - 2y

Hint:

This problem type is a straightforward application of partial differentiation rules. Remember that when differentiating with respect to one variable, all other variables are treated as constants.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a function \(z\) of two variables, \(x\) and \(y\), and we need to evaluate an expression involving its partial derivatives.

Step 2: Key Formula or Approach:
1. Find the partial derivative of \(z\) with respect to \(x\), \(\frac{\partial z}{\partial x}\), by treating \(y\) as a constant.
2. Find the partial derivative of \(z\) with respect to \(y\), \(\frac{\partial z}{\partial y}\), by treating \(x\) as a constant.
3. Substitute these derivatives into the given expression and simplify.

Step 3: Detailed Explanation:
The given function is \(z = x^2 - y^2\).
1. Calculate the partial derivative with respect to \(x\):
\[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x - 0 = 2x \]
2. Calculate the partial derivative with respect to \(y\):
\[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = 0 - 2y = -2y \]
3. Now, substitute these results into the given expression:
\[ \frac{1}{x} \frac{\partial z}{\partial x} + \frac{1}{y} \frac{\partial z}{\partial y} = \frac{1}{x} (2x) + \frac{1}{y} (-2y) \]
Simplify the expression:
\[ = 2 - 2 = 0 \]

Step 4: Final Answer:
The value of the expression is 0.