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$\begingroup$

For
$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} &\text{if }(x,y)\neq (0,0);\\ 0 &\text{if }(x,y)=(0,0).\end{cases}$

I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.

Taking the partial derivative freehand I get
$\frac{y(x^2+y^2)-xy(2x)}{(x^2+y^2)^2}= \frac{y(y^2-x^2)}{(x^2+y^2)^2}.$

But using the limits I get $\lim_{x\to c}\frac{f(x,y)-f(c,y)}{x-c} = y(y^2-c^2).$

I'm not sure what I'm doing wrong...

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    I was working really hard at trying to type up how I computed the limit with the math code and I figured out my mistake. I was missing part of my denominator. Thank you2010-11-24

2 Answers 2

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I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have: \begin{align*} \lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\ &= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\ &= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c)(x^2+y^2)(c^2+y^2)}\\ &= \lim{x\to c}\frac{y^3-xyc}{(x^2+y^2)(c^2+y^2)}\\ &= \frac{y(y^2-c^2)}{(c^2+y^2)^2}. \end{align*}

So, how did you get what you claim you got?

Added. For the partial at $(x,y)=(0,0)$, you have: $\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{0 - 0}{x} = 0.$

Since the limits exist, the partial derivative exists at each point.

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    @JimJones: The limit above works for every point other than $c=0$, $y=0$. Since the limit exists, the partial derivative exists at those points. For $y=0$ and $c=0$, you would need to do that limit. But in any case, this only gives the *partial* derivative, and even existence of both partials does not, in and of itself, imply that the derivative of the 2-variable function exists.2010-11-24
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You have an error in the calculation of the limit (see Arturo's remark about the denominator). When you correct it, you'll get the expression you calculated freehand, with $c$ substituted for $x$.