To find $f_x$ and $f_y$ of $f(x,y)=\frac{x \sqrt{x^2+y^2}}{|x|}$

Question

Let $$ f(x,y)=\frac{x \sqrt{x^2+y^2}}{|x|} $$ $$F = 0 $$$$at x=0$$ I need to find value of partial of $f$ wrt $x$ and $y$ at the origin. I have used definition and found that partials dont exist as absolute value symbols screws whole thing up . but there is answer (WELL DEFINED) in textbook and that is 0 .can any1 help me out pls

Answer 1

To find $f_x$ at $(0,0)$ one first puts $y=0$ in the formula for $f$, then takes the derivative of that at $x=0$. On putting $y=0$ we get $f=x \sqrt{x^2}/|x|$ which since $\sqrt{x^2}=|x|$ for all $x$ simplifies to $f=x.$ [One needs to put $f=0$ at $(0,0)$ here since given formula undefined/] But the derivative of that at $x=0$ is $1$ rather than the $0$ you have quoted.

Note: The given formula is not defined at any point $(0,y).$ If one defines it at those points as $0$ (as suggested in the recent edit of the question) then the above gives the partial $f_x(0,0)$ as $1$. Then also since $f$ is identically $0$ on the $y$ axis, one also has $f_y(0,0)=0.$

Answer 2

$$f(x,y)=\text{sgn}(x)\sqrt{x^2+y^2}.$$ $$\frac\partial{\partial x} f(x,y)=\text{sgn}(x)\frac x{\sqrt{x^2+y^2}},$$ $$\frac\partial{\partial y} f(x,y)=\text{sgn}(x)\frac y{\sqrt{x^2+y^2}}.$$

In polar coordinates,

$$\frac\partial{\partial x} f(x,y)=\text{sgn}(r\cos\theta)\cos\theta,$$ $$\frac\partial{\partial y} f(x,y)=\text{sgn}(r\cos\theta)\sin\theta.$$

The limits are not defined at the origin.