Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description hereenter image description here

$$\begin{align*} &f(x,y) = \frac{y}{x^2+y^2}\\ &f_{xx} = \frac{∂}{∂x}\left(-\frac{2xy}{(x^2+y^2)^2}\right)=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}\\ &f_{yy} = \frac{∂}{∂x}\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)=\frac{2y(y^2-3x^2)}{(x^2+y^2)^3} \end{align*}$$

$a$ and $b$ must be $(0,0)$ or at least one of them has to be $0$ to be critical points of the function

if the point is $(0,0)$, $f_{xx}$ or $f_{yy}$ will be undefined and finding the point $(?,0)$ for which $f_x$ or $f_y$ would $= 0$ would be mighty hard.

Are we unable to find the extrema of this function using the second partial test?

share|cite|improve this question
up vote 3 down vote accepted

First, you’ve some errors in your third line of calculations. The correct partials are:

$$\begin{align*} &f_x(x,y)=-\frac{2xy}{(x^2+y^2)^2}\\ &f_y(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\\ &f_{xx}(x,y)=\frac{2y(3x^2-y^2)}{(x^2+y^2)^3}\\ &f_{xy}(x,y)=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3}\\ &f_{yy}(x,y)=\frac{3y(y^2-2x^2)}{(x^2+y^2)^3} \end{align*}$$

Clearly $f_x(x,y)=0$ only if $xy=0$, i.e., at least one of $x$ and $y$ is $0$. Similarly, $f_y(x,y)=0$ only if $0=x^2-y^2=(x-y)(x+y)$, i.e., only if $y=x$ or $y=-x$. The only way to satisfy both of these conditions is to have $x=y=0$, and the function and its partial derivatives aren’t defined at $(0,0)$. Thus, you’re quite right: there is no point at which the second derivative test applies.

Rewriting the function in polar coordinates as


may help to explain what’s going on. As we travel around the circle $C_r$ of radius $r$ centred at the origin, the function value is $0$ where $C_r$ crosses the $x$-axis, reaches a maximum of $\frac1r$ where $C_r$ crosses the positive $y$-axis, and reaches a minimum of $-\frac1r$ where $C_r$ crosses the negative $x$-axis. But that high point with value $\frac1r$ on $C_r$ can’t be a local maximum of the function, because the value of the function gets larger as you move down the $y$-axis towards the origin: $\frac1r$ increases as $r$ decreases. Similarly, the low point on $C_r$ can’t be a local minimum of $f$, because $-\frac1r$ gets smaller (more negative) as you move up the negative $y$-axis and $r$ decreases.

share|cite|improve this answer
So, taking from the theroem, would we say "f does not have continuous second partial derivatives anywhere, thus the second partial test does not apply" ? – Matt Dec 15 '11 at 21:10
@Matt: That isn’t the problem: the second partials are continuous except at $(0,0)$, where they’re not defined. The problem is that there is no point where the first partials are both $0$, so there’s no point that is even a candidate to be a local maximum or minimum. – Brian M. Scott Dec 15 '11 at 21:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.