Question:medium

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

Show Hint

In partial differentiation, treat other variables as constants.
  • \(1\)
  • \(2x+2y\)
  • \(0\)
  • \(2x-2y\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
This problem involves calculating partial derivatives of a function of two variables and then substituting them into a given expression. Partial differentiation means differentiating with respect to one variable while treating all other variables as constants.
Step 2: Key Formula or Approach:
1. Calculate the partial derivative of $z$ with respect to $x$, $\frac{\partial z}{\partial x}$. 2. Calculate the partial derivative of $z$ with respect to $y$, $\frac{\partial z}{\partial y}$. 3. Substitute these derivatives into the given expression and simplify.
Step 3: Detailed Explanation:
The given function is $z = x^2 - y^2$. 1. Find the partial derivative with respect to $x$ (treating $y$ as a constant): \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x - 0 = 2x \] 2. Find the partial derivative with respect to $y$ (treating $x$ as a constant): \[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = 0 - 2y = -2y \] 3. Substitute these results into the expression $\frac{1}{x} \frac{\partial z}{\partial x} + \frac{1}{y} \frac{\partial z}{\partial y}$: \[ \frac{1}{x}(2x) + \frac{1}{y}(-2y) \] \[ = 2 - 2 = 0 \] Step 4: Final Answer:
The value of the expression is 0. Therefore, option (C) is correct.
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