Question

$I$f $f(x,y)=\frac{1}{x^2}+\frac{1}{xy}+\frac{\log x-\log y}{x^2+y^2}$, then $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}$ is equal to

• A. $0$
• B. $f$
• C. $2f$
• D. $-2f$

Answer 1

The correct option is D $-2f$

Given,

$f(x,y)=\frac{1}{x^2}+\frac{1}{xy}+\frac{\log x-\log y}{x^2+y^2}$

So, by partially differentiating the function w.r.t x and multiplying with x, we get,

$\Rightarrow x\frac{\partial f}{\partial x}=\frac{-2}{x^2}-\frac{1}{xy}+\frac{(x^2+y^2)-(\log x-\log y)\left(2x^2\right)}{{(x^2+y^2)}^2}$

similiarly, if we partially differentiate w.r.t y and multiply it with y, we get,

$\Rightarrow y\frac{\partial f}{\partial y}=-\frac{1}{xy}+\frac{-(x^2+y^2)-(\log x-\log y)\left(2y^2\right)}{{(x^2+y^2)}^2}$

Adding both the equations we get,

$\Rightarrow x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=-\frac{2}{x^2}-\frac{2}{xy}-\frac{(\log x-\log y)(2x^2+2y^2)}{{(x^2+y^2)}^2}=\frac{2}{x^2}-\frac{2}{xy}-\frac{2(\log x-\log y)}{{(x^2+y^2)}^2}=-2f$