Question

If $v=log\frac{x^2+y^2}{x+y}$ prove that $x\cdot \frac{\partial v}{\partial x}+y\cdot \frac{\partial v}{\partial y}=1$

Answer 1

Given,

$v=log\frac{x^2+y^2}{x+y}$

$v=log(x^2+y^2)-\log (x+y)$....(1)

Differentiating equation (1) with respect to x

Now, $\frac{\partial v}{\partial x}=\frac{2x}{x^2+y^2}-\frac{1}{x+y}$.......(2).

And Differentiating equation (2) with respect to y

$\frac{\partial v}{\partial y}=\frac{2y}{x^2+y^2}-\frac{1}{x+y}$.......(3).

$\therefore$ $x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}=\frac{2(x^2+y^2)}{x^2+y^2}-\frac{x+y}{x+y}=2-1=1$. [ Adding equation (2) and (3)]