Question

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

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

Hint:

In partial differentiation, while differentiating with respect to one variable, treat the other variable as constant.

Answer 1

The Correct Option is C

We are given: \[ z=x^2-y^2. \] We need to find: \[ \frac{1}{x}\frac{\partial z}{\partial x}+\frac{1}{y}\frac{\partial z}{\partial y}. \] First find partial derivative with respect to \(x\). While differentiating with respect to \(x\), treat \(y\) as constant. \[ \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(x^2-y^2). \] \[ \frac{\partial z}{\partial x}=2x. \] Now find partial derivative with respect to \(y\). While differentiating with respect to \(y\), treat \(x\) as constant. \[ \frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(x^2-y^2). \] \[ \frac{\partial z}{\partial y}=-2y. \] Now substitute: \[ \frac{1}{x}\frac{\partial z}{\partial x}+\frac{1}{y}\frac{\partial z}{\partial y} = \frac{1}{x}(2x)+\frac{1}{y}(-2y). \] \[ =2-2. \] \[ =0. \] Hence, the required value is \[ 0. \]