Question

If $\mathrm{u}=\frac{(\mathrm{x}+\mathrm{y})}{(\mathrm{x}-\mathrm{y})}$, then$\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=$

• A. $\frac{1}{(\mathrm{x}-\mathrm{y})}$
• B. $\frac{2}{\left(\mathrm{x}-\mathrm{y}\right)}$
• C. $\frac{1}{{\left(x-y\right)}^2}$
• D. $\frac{2}{{(x-y)}^2}$

Answer 1

The correct option is B

$\frac{2}{\left(\mathrm{x}-\mathrm{y}\right)}$

Explanation for the correct option:

Finding the value of the expression:

Given equation,

$u=\frac{\left(x+y\right)}{\left(x-y\right)}$

Partially differentiating the above equation with respect to $x$ and $y$ respectively.

$$\begin{gathered} \frac{\partial u}{\partial x}=\frac{\left[(x-y)–(x+y)\right]}{{(x-y)}^2} \\ \Rightarrow \frac{\partial u}{\partial x}=\frac{-2y}{{(x-y)}^2} \\ \frac{\partial u}{\partial y}=\frac{\left[\left(x-y\right)+\left(x+y\right)\right]}{{(x-y)}^2} \\ \Rightarrow \frac{\partial u}{\partial y}=\frac{2x}{{(x-y)}^2} \end{gathered}$$

Substitute these values in the finding expression:

$\begin{array}{rcl}\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}& =& \left[\frac{-2y}{{(x-y)}^2}\right]+\left[\frac{2x}{{\left(x-y\right)}^2}\right]\\ & =& \frac{2\left(x-y\right)}{{(x-y)}^2}\\ & =& \frac{2}{(x-y)}\end{array}$

Hence, the correct option is (B).